THE  LIBRARY 

OF 

THE  UNIVERSITY 
OF  CALIFORNIA 

LOS  ANGELES 


SOUTHERN  BRANCH' 

UNIVERSITY  OF  CALIFORNIA 

LIBRARY, 

,  CALIF. 


LIBRARY 

OF 

PRACTICAL  ELECTRICITY 


VOLUME  I 


JM  Book  Co.  Jn& 


PUBLISHERS     OP     BOOKS      F  O  P-_/ 

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PRACTICAL  MATHEMATICS 

FOR 

HOME  STUDY 


BEING  THE  ESSENTIALS  OF 

ARITHMETIC,  GEOMETRY, 
ALGEBRA  AND  TRIGONOMETRY 


BY 

CLAUDE  IRWIN  PALMER 

ASSOCIATE    PROFESSOR   OF   MATHEMATICS   ARMOUR  INSTITUTE    OF  TECHNOLOGY 


FIRST  EDITION 
TWELFTH  IMPRESSION 


McGRAW-HILL  BOOK  COMPANY,  INC. 
NEW  YORK:  370  SEVENTH  AVENUE 

LONDON:  6  &  8  BOUVERIE  ST.,  E.  C.  4 

r*  n  4>  ***  r* 

5  u  &  i  ;> 


COPYRIGHT,  1919,  BY  THE 
McGRAW-HiLL  BOOK  COMPANY,  INC. 

PRINTED    IN    THE   UNITED   STATES   OF   AMERICA 


TMK    M  A  I-  *,  K    J'KKHH    VOMK    !•  A 


College 
Library 


PREFACE 

During  the  past  fifteen  years  the  author  has  taught  classes 
in  practical  mathematics  in  the  evening  school  at  the  Armour 
Institute  of  Technology,  Chicago.  These  classes  have  been 

H  composed  of  men  engaged  in  practical  pursuits  of  various 

^  kinds.     The  needs  of  these  men  have  been  carefully  studied; 

:v-  and,  so  far  as  possible,  those  mathematical  subjects  of  interest 
to  them  have  been  taken  up.  The  matter  presented  to  the 
classes  has  necessarily  been  of  an  intensely  practical  nature. 

^  This  has  been  worked  over  and  arranged  in  a  form  that  was 

•^ 

thought  most  suitable  for  class  use;  and  was  printed  in  Pal- 
Ui  mer's  Practical  Mathematics,  four  volumes,  in  1912  and  ap- 
v  peared  in  a  revised  edition  in  1918. 

The  four  volume  edition  has  been  used  by  thousands  of 

men  for  home  study.     It  is  to  meet  the  needs  especially  of 

such  men  that  this  one  volume  edition  has  been  made.     The 

vj  subject  matter  includes  all  that  is  in  the  four  volumes;  and  to 

*  this  has  been  added  a  few  new  topics  together  with  many  solu- 

3  tions  of  exercises,  and  suggestions  that  make  the  text  more 

I  suitable  for  home  study.     It  is  hoped  that  it  will  find  a  place 

^  in  the  library  of  the  man  who  applies  elementary  mathematics. 

v 

5  and  who  wishes  occasionally  to  brush  up  his  mathematics. 

Usually  when  the  practical  man  appreciates  the  fact  for 
himself  that  mathematics  is  a  powerful  tool  that  he  must  be 
able  to  use  in  performing  his  work,  he  finds  that  even  the 
arithmetic  that  he  learned  at  school  has  left  him.  A  student 
of  this  kind  is  discouraged  if  required  to  pursue  the  study  of 
mathematics  in  the  ordinary  text-books. 

This  work  has  been  written  for  the  adult.  The  endeavor 
has  been  to  make  the  student  feel  that  he  is  in  actual  touch 
with  real  things.  The  intention  has  been  to  lay  as  broad  a 
foundation  as  is  consistent  with  the  scope  of  the  work. 

The  nearly  3000  drill  exercises  and  problems  are,  in  most 


vi  PREFACE 

cases,  new.  Many  of  them  are  adapted  from  engineering  and 
trade  journals,  from  handbooks  of  various  kinds,  and  from 
treatises  on  the  steel  square  and  other  mechanical  devices; 
other  problems  are  from  the  author's  experience;  and  a  largo 
number  of  the  specially  practical  problems  were  proposed  by 
members  of  the  classes  pursuing  the  course  during  its  growth. 

Much  information  on  various  matters  to  which  mathe- 
matics is  applied,  is  incidentally  given  in  the  problems.  Many 
devices  and  methods  used  by  the  practical  man  are  given. 
Care  has  been  taken  to  make  these  true  to  practice;  but,  in  so 
wide  a  range  of  matter,  there  are  undoubtedly  errors.  It  is 
thought  that  the  answers  to  the  exercises  are  given  to  a  reason- 
able degree  of  accuracy.  It  is  hoped  that  the  volume,  as  a 
whole,  will  not  be  found  unmathematical. 

The  main  features  of  Part  I  are  the  concise  treatment  of 
various  subjects  in  arithmetic  and  their  applications,  checks 
of  processes,  degree  of  accuracy  possible  in  solutions,  and 
contracted  processes. 

In  Part  II,  the  endeavor  has  been  to  state  definitions  so  as 
to  give  a  clear  idea  of  the  term  or  object  defined,  and  yet  not 
to  be  too  technical.  Wherever  possible,  the  attempt  is  made 
to  discuss  a  fact  or  principle  of  geometry  in  such  a  way  that 
its  reasonableness  will  be  apparent.  While  the  subjects  are 
treated  in  the  mathematical  order,  many  applications  are 
given  under  separate  headings.  Such  are  brickwork,  lumber, 
the  steel  square,  screw  threads,  circular  mils,  belt  pulleys,  and 
gear  wheels. 

In  Part  III,  the  intention  is  to  give  sufficient  drill  in  algebra 
for  one  who  wishes  to  make  direct  applications  to  practical 
problems.  Much  attention  is  given  to  formulas  and  their 
transformations.  The  equation  is  applied  to  many  practical 
problems.  Graphical  methods  are  considered,  and  many 
articles  on  special  subjects  are  given. 

In  Part  IV,  the  intention  is  to  give  sufficient  work  in 
logarithms  to  secure  a  fair  degree  of  skill  in  computations. 
In  trigonometry,  those  parts  are  emphasized  that  may  be 
applied  directly  to  practical  problems;  while  the  portions 
chiefly  necessary  as  an  aid  in  the  study  of  more  advanced 
mathematical  subjects,  are  either  treated  very  slightly  or 


PREFACE  vii 

omitted.     Many  applications  are  given.     The  tables  are  given 
to  four  decimal  places. 

The  author  wishes  to  acknowledge  his  great  indebtedness 
to  the  more  than  1000  men  who  made  up  his  classes  during 
the  growth  of  this  work,  and  to  the  hundreds  of  men  from 
various  parts  of  the  country  who  have  offered  kindly  criti- 
cisms and  suggestions;  for,  without  their  help  and  sympathy, 
the  present  results  would  have  been  impossible. 

Because  of  the  remarkable  success  of  the  previous  editions, 
it  is  with  the  greatest  pleasure  that  this  special  edition  is  sub- 
mitted to  our  practical  men. 

C.  I.  PALMER. 

CHICAGO, 
June,  1919. 


A  WORD  WITH  THE   STUDENT 

One  of  the  lessons  of  the  Great  War  and  the  strenuous 
efforts  necessary  to  carry  it  on,  has  been  to  bring  forcibly  to 
our  minds  the  great  usefulness  of  mathematics.  The  war 
activities  have  exhibited  the  extensive  mathematical  needs  of 
those  who  aim  to  render  the  most  efficient  service  under  the 
most  trying  circumstances.  The  young  men  of  the  country 
realize  the  need  for  a  working  knowledge  of  mathematics, 
and  see  clearly  that  the  need  so  emphasized  by  the  war  con- 
ditions is  being  carried  over  into  the  days  of  peace  and  into 
the  period  of  great  industrial  activity  that  is  sure  to  follow. 

This  volume  being  entitled  Practical  Mathematics  does 
not  mean  that  all  exercises  are  such  as  would  be  called  prac- 
tical. It  means  that,  in  the  main,  the  exercises,  outside  of 
those  intended  for  pure  drill,  are  such  as  may  arise  in  some 
practical  field  of  work.  The  endeavor  has  been  to  utilize 
the  material  afforded  by  the  shops  and  the  laboratories  as 
well  as  in  the  trades  and  in  engineering. 

The  practical  man  realizes  that,  for  him,  mathematics  is 
a  chest  of  tools  together  with  many  more  or  less  complicated 
pieces  of  machinery  that  he  may  use  to  accomplish  his  purpose. 
To  apply  mathematics,  then,  he  must  be  able  to  run  its 
machinery  not  only  accurately  but  speedily.  To  do  this  a 
great  deal  of  work  must  be  done  in  the  arithmetical  processes 
themselves.  The  student  must  drill  himself  on  the  funda- 
mental operations — addition,  subtraction,  multiplication,  and 
division — both  in  whole  numbers  and  in  fractions,  until  the 
processes  become  to  a  large  degree  mechanical.  That  is, 
he  should  be  able  to  do  these  operations  with  but  little 
expenditure  of  mental  energy.  This  drill  is  best  gained  by 
doing  many  exercises  especially  set  for  this  purpose.  Each 
student  who  is  studying  alone,  that  is,  without  being  in  a 
class,  must  of  necessity  be  his  own  judge  as  to  how  much 
drill  he  needs.  For  most  people  such  drill  is  tedious  and 


x  PRACTICAL  MATHEMATICS 

uninteresting,  and  it  requires  a  strong  will  to  force  oneself  to 
do  the  proper  amount  of  such  work. 

That  these  ideas  are  not  new  is  evident  from  the  following 
quotation  in  quaint  old  English,  taken  from  an  arithmetic 
printed  more  than  two  hundred  years  ago:  "Therefore, 
Courteous  Reader,  if  thou  intendest  to  be  a  Proficient  in  the 
Mathematicks,  begin  cheerfully,  proceed  gradually,  and  with 
Resolution,  and  the  end  will  crown  thy  Endeavours  with 
Success;  and  be  not  so  slothfully  Studious,  as  at  every  Diffi- 
culty thou  meetest  withal  to  cry,  Ne  plus  ultra,  for  Pains 
and  Diligence  will  overcome  the  greatest  Difficulty:  To 
conclude,  That  thou  may'st  so  read  as  to  understand,  and  so 
understand,  as  to  become  a  Proficient,  is  the  hearty  cfesire 
of  him  who  wisheth  thy  Welfare,  and  the  Progress  of  Arts. 
From  my  School  at  St.  George's  Church  in  Southwork,  October 
27,  1684." 


CONTENTS 

PAGE 

PREFACE , v 

A  WORD  WITH  THE   STUDENT ix 

PART  I 

ARITHMETIC 

CHAPTER  I 

PRELIMINARY  WORK  AND  REVIEW 

ART.  PAGE 

1.  Language  of  mathematics 1 

2.  How  to  attack  a  problem 3 

3.  Definitions 4 

4.  Rules  for  finding  divisor  of  numbers 7 

5.  Relative  importance  of  signs  of  operation 2 

6.  Cancellation 9 

7.  Applying  Rules 2 


CHAPTER  II 

COMMON  FRACTIONS 

9.  Definitions 12 

10.  Mixed  number 12 

11.  Proper  and  improper  fractions 12 

12.  Comparison  of  fractions 13 

13.  Remarks 13 

14.  Principles 14 

15.  Reduction  of  a  whole  or  a  mixed  number  to  an  improper  frac- 

tion    14 

16.  Reduction  of  an  improper  fraction  to  a  whole  or  mixed  number .  14 

17.  Reduction  of  fractions  to  lowest  terms 15 

18.  Reduction  of  several  fractions  to  fractions  having  the  same 

denominator 15 

19.  Least  common  denominator 16 

20.  Addition  of  fractions 17 

21.  Subtraction  of  fractions 19 

22.  Resultants 21 

xi 


xii  CONTENTS 

ART.  PACK 

23.  Multiplication  of  fraction  and  integer 21 

24.  Multiplication  of  a  fraction  by  a  fraction 22 

25.  Multiplication  of  mixed  numbers  and  integers 23 

26.  Division  of  a  fraction  by  an  integer 26 

27.  Division  by  a  fraction 27 

28.  Special  methods  in  division 27 

29.  Pitch  and  lead  of  screw  threads 31 

30.  The  micrometer 32 

32.  Screw  gearing 33 

CHAPTER  III 

DECIMAL  FRACTIONS 

32.  Definition 38 

33.  Reading  numbers 39 

34.  Reduction  of  a  common  fraction  to  a  decimal  fraction  ....  40 

35.  Decimal  fraction  to  common  fraction 40 

36.  Addition  of  decimals 41 

37.  Subtraction  of  decimals 41 

38.  Multiplication  of  decimals 42 

39.  Division  of  decimals 42 

40.  Accuracy  of  results 43 

41.  Proportions  of  machines  screw  heads.     A.  S.  M.  E.  standard.  49 

CHAPTER  IV 
SHORT  METHODS  AND  CHECKS 

42.  Contracted  methods  and  approximate  results 52 

43.  Other  methods 53 

44.  Checking 55 

CHAPTER  V 

WEIGHTS  AND  MEASURES 

45.  English  system 58 

46.  The  Metric  system 62 

47.  Measure  of  length.     The  Meter 63 

48.  Legal  units 63 

49.  Measure  of  surface 64 

50.  Measures  of  volume.     Cubic  and  capacity  measures 64 

51.  Measures  of  mass 64 

52.  Tables  and  terms  used 65 

53.  Equivalents 67 

54.  Simplicity  of  the  metric  system 68 

55.  Relations  of  the  units 69 

56.  Changing  from  English  to  metric  or  from  metric  to  English 

systems 69 


CONTENTS  xiii 

CHAPTER  VI 

PKHCENTAGK  AND  APPLICATIONS 
ART.  PA«K 

57.  Per  cents  as  fractions 73 

58.  Cases 73 

59.  Rules  and  formulas 74 

60.  Solutions       75 

61.  Applications 76 

62.  Averages  and  per  cent  of  error 77 

63.  List  prices  and  discounts 78 

64.  Interest  . 83 

CHAPTER  VII 
RATIO  AND  PROPORTION 

65.  Ratio 85 

66.  Proportion 86 

67.  Measuring  heights 89 

68.  The  lever 91 

69.  Hydraulic  machines 93 

CHAPTER  VIII 
DENSITY  AND  SPECIFIC  GRAVITY 

70.  Density 94 

71.  Specific  gravity 94 

72.  Standards 95 

73.  Use 94 

CHAPTER  IX 
POWER  AND  ROOTS 

74.  Powers 98 

75.  Exponent  of  a  power 98 

76.  Squares,  cubes,  involution 98 

77.  Roots 99 

78.  Radical  sign  and  index  of  root 99 

79.  Square  root 99 

80.  Process  for  the  square  root  of  a  perfect  square 100 

81.  Square  root  of  a  number  containing  a  decimal 102 

82.  Roots  not  exact 102 

83.  Root  of  a  common  fraction 103 

84.  Short  methods 103 

85.  Rule  for  square  root 105 

86.  Cube  root 105 

87.  Similar  figures 106 


xiv  CONTENTS 

PART  II 
GEOMETRY 
CHAPTER  X 

PLANE  SURFACES.    LINES  AND  ANGLES 
ART  .  PACK 

89.  Definitions 109 

90.  Angles Ill 

91.  Polygons 112 

92.  Concerning  triangles 113 

93.  Concerning  quadrilaterals 114 

94.  The  rectangle 115 

95.  The   parallelogram     ...  .115 

96.  Formulas      .116 

97.  The  triangle 116 

98.  Area  of  a  triangle  when  the  three  sides  only  are  given  .    .    .    .117 

99.  Area  of  trapezoid 118 

100.  Measuring  lumber 121 

101.  Estimations 122 

102.  Shingles 122 

CHAPTER  XI 
TRIANGLES 

103.  A  right  triangle 126 

104.  Similar  triangles 130 

105.  Tapers 131 

106.  Turning 132 

107.  The  steel  square 134 

108.  Rafters  and  roofs 135 

109.  Uses  of  the  square 136 

110.  Isosceles  and  equilateral  triangles 139 

111.  The  isosceles  triangle 139 

112.  The  equilateral  triangle 139 

113.  The  regular  hexagon 141 

114.  Screwthreads 143 

115.  Sharp  V-thread 144 

116.  United  States  standard  thread 144 

CHAPTER  XII 

CIRCLES 

118.  Definitions 147 

119.  Properties  of  the  circle 149 


CONTENTS  xv 

ART.  PAGE 

120.  The  segment 150 

121.  Relations  between  the  diameter,  radius,  and  circumference .    .  151 

122.  Area  of  the  circle 151 

123.  Area  of  a  ring 153 

124.  Area  of  a  sector 153 

125.  Area  of  a  segment 154 

126.  The  ellipse 155 

127.  Regular  polygons  and  circles 166 

128.  Rules 169 

129.  Feed 170 

130.  Cutting  speeds 170 

131.  Belt  pulleys  and  gear  wheels 171 

132.  The  circular  mil 175 

133.  The  square  mil 176 

CHAPTER  XIII 

GRAPHICAL  METHODS 

134.  Units 177 

135.  Circular  measure,  radian 177 

136.  The  protractor 178 

137.  To  measure  an  angle  with  protractor 179 

138.  To  lay  off  an  angle  with  a  pi  otr actor 179 

139.  Angles  and  chords 180 

140.  To  find  a  chord  length  from  the  table 180 

141.  To  lay  off  an  angle 180 

142.  To  measure  an  angle .  181 

143.  Drawing  to  scale 181 

144.  To  construct  a  triangle  having  given  two  sides  and  the  angle 

between  these  sides 181 

145.  To  construct  a  triangle  when  given  two  angles  and  the  side 

between  these  angles 182 

146.  To  construct  a  triangle  when  the  three  sides  are  given.    .    .    .    182 

147.  Areas  found  by  the  use  of  squared  paper 183 

148.  Other  methods  for  approximating  areas 184 

149.  To  divide  a  line  of  any  length  into  a  given  number  of  equal  parts  186 

150.  To  cut  off  the  corners  of  a  square  so  as  to  form  a  regular  octagon  187 

151.  To  divide  a  given  circle  into  any  number  of  equal  parts  by  con- 

centric circles 187 

152.  To  inscribe  regular  polygons 188 

153.  To  draw  the  arc  of  a  segment  when  the  chord  and  the  height 

of  the  segment  are  given 189 

154.  To  find  the  radius  of  a  circle  when  only  a  part  of  the  circumfer- 

ence is  known 189 

155.  How  to  cut  a  strikeboard  to  a  circular  arc 190 

156.  The  vernier 191 

157.  Micrometer  with  vernier  .  .192 


xvi  CONTENTS 

CHAPTER  XIV 

PltlBMS 
ART.  PACK 

158.  Definitions 194 

159.  Surfaces 195 

160.  Volumes 196 

161.  Terms 200 

162.  Estimating  cost  of  stonework 200 

163.  Brick 201 

164.  Estimating  number,  and  cost  of  brickwork 201 


CHAPTER  XV 

CYLINDERS 

165.  Definitions 203 

166.  Area  and  volume 203 

167.  The  hollow  cylinder 204 


CHAPTER  XVI 

PYRAMIDS,  CONES,  AND  FRUSTUMS 

168.  Pyramid 214 

169.  Cone 214 

170.  Frustum 215 

171.  Areas 216 

172.  Volumes  .  .  216 


CHAPTER  XVII 

THE  SPHERE 

173.  Definitions 220 

174.  Area 220 

175.  Volume 221 

176.  Zone  and  segment  of  sphere 221 


CHAPTER  XVIII 

VARIOUS  OTHER  SOLIDS 

177.  Anchor  ring 226 

178.  Prtsmatoids    .  .  227 


CONTENTS  xvii 

PART  III 

ALGEBRA 

CHAPTER  XIX 

NOTATION  AND  DEFINITIONS 

ART.  PAGE 

179.  General  remarks 231 

180.  Definite  numbers .231 

181.  General  numbers 231 

182.  Signs 232 

183.  Algebraic  expression 233 

184.  Coefficient 233 

185.  Power,  exponent 233 

186.  Terms 234 

187.  Remarks 234 

CHAPTER  XX 
FORMULAS  AND  TRANSLATIONS 

188.  Subject  matter 237 

189.  The  slide  rules 237 

190.  Evaluation  of  algebraic  expressions 238 

CHAPTER  XXI 
POSITIVE  AND  NEGATIVE  NUMBERS 

191.  Meaning  of  negative  numbers 244 

192.  Need  of  negative  number 244 

193.  Representation  of  negative  and  positive  numbers 245 

194.  Definitions 245 

195.  Remarks  on  numbers 246 

CHAPTER  XXII 

ADDITION  AND  SUBTRACTION 

196.  Definitions 248 

197.  Addition  of  algebraic  numbers 248 

198.  Principles 249 

199.  Subtraction  of  algebraic  numbers 249 

200.  Addition  and  subtraction  of  literal  algebraic  expressions  .    .    .  250 

201.  Polynomials 251 

202.  Test  or  proof  of  results 251 

203.  Terms  with  unlike  coefficients 253 

204.  Signs  of  grouping 254 

205.  Insertion  of  signs  of  grouping 255 


xviii  CONTENTS 

CHAPTER  XXIir 

EQUATIONS 

ART.  PAGE 

206.  Definitions 257 

207.  The  equation 258 

208.  Solution  of  equations 268 

209.  Axioms 259 

210.  Testing  the  equation 260 

211.  The  equation  in  solving  problems 261 

CHAPTER  XXIV 
MULTIPLICATION 

212.  Fundamental  ideas 265 

213.  Rules 266 

214.  Concrete  illustration 266 

215.  Continued  products 267 

216.  Law  of  exponents 268 

217.  To  multiply  a  monomial  by  a  monomial 268 

218.  To  multiply  a  polynomial  by  monomial 269 

219.  To  multiply  a  polynomial  by  a  polynomial 269 

220'  Test 270 

221.  Representation  of  products 272 

222.  Approximate  products 273 


CHAPTER  XXV 
DIVISION,  SPECIAL  PRODUCTS,  AND  FACTORS 

223.  Division 275 

224.  Division  of  one  monomial  by  another 275 

225.  Test . .    .  276 

226.  Division  of  a  polynomial  by  a  monomial 277 

227.  Factors  of  a  polynomial  when  one  factor  is  a  monomial    .    .    .  277 

228.  Squares  and  square  roots  of  monomials 278 

229.  The  square  of  a  binomial .  279 

230.  Factors  of  a  trinomial  square 280 

231.  The  product  of  the  sum  of  two  numbers  by  the  difference  of  the 

same  two  numbers 280 

232.  Factors  of  the  difference  of  two  squares 281 

233.  The  product  of  two  binomials  having  one  common  term  .    .    .  282 

234.  To  factor  a  trinomial  into  two  binomials  with  one  common  term  283 

235.  Other  forms    .  .  283 


CONTENTS  xix 
CHAPTER  XXVI 

EQUATIONS 

236.  Solution    . 285 

237.  Equations  solved  by  aid  of  factoring 287 

238.  Formulas 290 

CHAPTER  XXVII 

FRACTIONS 

240.  Reduction  of  a  fraction  to  its  lowest  terms 293 

241.  Reduction  of  fractions  to  common  denominators 294 

242.  Lowest  common  multiple .  295 

243.  Fractions  having  a  L.  C.  D 295 

244.  Addition  and  subtraction  of  fractions 297 

245.  Multiplication  of  fractions 298 

246.  Division  of  fractions.  299 


CHAPTER  XXVIII 
EQUATIONS  AND  FORMULAS 

247.  Subject  matter 303 

248.  Order  of  procedure 303 

249.  Clearing  of  fractions 303 

250.  Thermometers 311 

251.  Horse-power 312 

252.  Relation  of  resistance,  electromotive  force,  and  current.    .    .  316 

253.  Resistance  of  conductors 317 


CHAPTER  XXIX 

EQUATIONS  WITH  MORE  THAN  ONE  UNKNOWN 

255.  Indeterminate    equations 319 

256.  Simultaneous  equations 319 

257.  Solution  of  independent  equations 320 

258.  Elimination  by  adding  or  subtracting 320 

259.  Elimination  by  substitution 321 

260.  Elimination  by  comparison 322 

261.  Suggestions 322 


XX  CONTENTS 

CHAPTER  XXX 

EXPONENTS,  POWKKS,  AND  ROOTS 

ART.  PACE 

262.  General  statement 327 

263.  Laws  of  exponents 327 

264.  Zero  exponent 329 

265.  Negative  exponent 329 

266.  Fractional  exponent 330 

267.  Exponents  used  in  writing  numbers 331 

CHAPTER  XXXI 

QUADRATIC  EQUATIONS 

268.  Definitions ...  332 

269.  Solution 332 

270.  Solution  by  factoring .    .  334 

272.  Completing  the  square 334 

273.  Solution  by  completing  the  square 335 

274.  Solution  of  the  affected  quadratic  equation  by  the  formula  .    .  337 

CHAPTER  XXXII 

VARIATION 

275.  General  statement 340 

276.  Constants  and  variables 340 

277.  Direct  variation 340 

278.  Mathematical  statement 341 

279.  Inverse  variation 342 

280.  Mathematical  statement 342 

281.  Joint  variation 343 

282.  Transverse  strength  of  wooden  beams 346 

283.  The  constant 347 

284.  Factor  of  safety 348 

CHAPTER  XXXIII 

GRAPHICS 

285.  The  graph 351 

286.  Definitions  and  terms  used 352 

287.  Plotting  points 354 

288.  Graph  of  an  equation 359 

289.  Simultaneous  equations 360 

290.  The  graph  of  an  equation  of  any  degree 361 

291.  Simpson's  Rule 363 

292.  The  average  ordinate  rule 364 

293.  Area  in  a  closed  curve 364 

294.  The  steam  indicator  diagram 365 


CONTENTS  xxi 

PART  IV 
LOGARITHMS  AND  TRIGONOMETRY 

CHAPTER  XXXIV 

LOGARITHMS 
ART.  PAGE 

295.  Uses 369 

296.  Exponents 369 

297.  Definitions  and  history 370 

298.  Notation 370 

299.  Illustrative  computations  by  means  of  exponents 372 

300.  Logarithms  of  any  number 373 

301.  Logarithms  to  the  base  10 373 

302.  Rules  for  determining  the  characteristic 374 

303.  The  mantissa 375 

304.  Tables 375 

305.  To  find  the  mantissa  of  a  number 376 

306.  Rules  for  finding  the  mantissa 377 

307.  Finding  the  logarithm  of  a  number 377 

308.  To  find  the  number  corresponding  to  a  logarithm 379 

309.  Rules  for  finding  the  number  corresponding  to  a  given  loga- 

rithm  380 

310.  To  find  the  product  of  two  or  more  factors  by  the  use  of  loga- 

rithms  381 

311.  To  find  the  quotient  of  two  numbers  by  logarithms 381 

312.  To  find  the  power  of  a  number  by  logarithms 383 

313.  To  find  the  root  of  a  number  by  logarithms 383 

314.  Computations  made  by  logarithms  only  approximate    ....   383 

315.  Natural  logarithms 384 

CHAPTER  XXXV 

INTRODUCTION,  ANGLES 

316.  Introductory 396 

317.  Angles 396 

318.  Location  of  angles,  quadrants 398 

319.  Measurement  of  angles 398 

320.  Relations  between  angle,  arc,  and  radius 400 

321.  Railroad  curves 400 

CHAPTER  XXXVI 
TRIGONOMETRIC  FUNCTIONS 

322.  Sine,  cosine,  and  tangent  of  an  acute  angle 404 

323.  Ratios  for  an  angle 404 

324.  General  form  for  ratios 405 

325.  Acute  angle  in  a  right  triangle 406 


xxh  CONTENTS 

ART.  PAc» 

326.  Relation  between  the  functions  of  an  angle  and  the  functions 

of  its  complement 407 

327.  Trigonometric  functions  by  construction  and  measurement .    .  408 

328.  Use  of  functions  in  constructing  angles 409 

329.  Values  of  functions  by  computation 410 

330.  Angles  in  other  quadrants 410 

331.  Angles  of  90°,  180°,  270°,  and  0° 411 

332.  Table  of  functions 412 

CHAPTER  XXXVII 
TABLES  AND  THEIR  USES 

333.  Nature  of  trigonometric  functions 414 

334.  Table  of  functions 414 

335.  To  find  the  function  of  an  angle  from  the  table 414 

336.  To  find  the  angle  corresponding  to  a  function 415 

337.  Evaluation  of  formulas 417 

CHAPTER  XXXVIII 
RIGHT  TRIANGLES 

339.  Solving 421 

340.  The  right  triangle 422 

341.  Directions  for  solving 422 

342.  Case  I.     Given  A  and  b,  A  and  a,  B  and  a,  or  B  and  6    ...  423 

343.  Directions  for  solution  of  triangles 424 

344.  Case  II.     Given  A  and  c  or  B  and  c 425 

345.  Case  III.     Given  c  and  a  or  c  and  6 425 

346.  Remark  on  inverse  functions 426 

347.  Case  IV.     Given  o  and  b 426 

348.  Orthogonal  projection 427 

349.  Vectors 427 

350.  Definitions 429 

35  J.  Widening  of  pavements  on  curves 438 

352.  Spirals 439 

CHAPTER  XXXIX 
RELATIONS  BETWEEN  RATIOS,  AND  PLOTTING 

353.  Relations  between  the  ratios  of  an  angle  and  the  ratios  of  its 

complement 442 

354.  Relations  between  the  ratios  of  an  angle  and  the  ratios  of  its 

supplement 442 

355.  Relations  between  ratios  of  an  angle  0  and  90°  +  0 443 

356.  Relations  between  the  ratios  of  an  angle  and  the  ratios  of  its 

negative 444 


CONTENTS  xxiii 

ART.  PAGE 

357.  Relations  between  the  ratios  of  any  angle 445 

358.  Plotting  the  sine  curve 447 

359.  Curves  for  cosine,  tangent,  cotangent,  secant,  and  cosecant.    .  449 
3CO.  Projections  of  a  point  having  circular  motion 449 

361.  Sine  curves  of  different  frequency  .    .    .    „ 452 

362.  Variation  in  the  amplitude  of  sine  curves 453 

CHAPTER  XL 

TRIGONOMETRIC  RATIOS  OF  MORE  THAN  ONE  ANGLE 

364.  Functions  of  the  sum  or  difference  of  two  angles 455 

365.  Functions  of  twice  an  angle  and  half  an  angle 456 

366.  Formulas  for  changing  products  to  sums  or  differences,    and 

sums  and  differences  to  products 457 

CHAPTER  XLI 

SOLUTION  OP  OBLIQUE  TRIANGLES 

367.  Cases 459 

368.  The  law  of  sines 459 

369.  The  law  of  cosines 460 

370.  Directions  for  solving 461 

371.  Case  I,  a  side  and  two  angles  given 461 

372.  Case  II,  two  sides  and  an  angle  opposite  one  of  them  given.    .  462 

373.  Case  III,  two  sides  and  the  angle  between  them  given  ....  464 

374.  Case  IV,  three  sides  given 464 

375.  Resultant  of  forces    .    .    .    .    , 466 

376.  Computation  of  a  resultant 467 

TABLES 

I.  SUMMARY  OF  FORMULAS 470 

II.  USEFUL  NUMBERS 474 

III.  DECIMAL  AND  FRACTIONAL  PARTS  OF  AN  INCH  ..'...-  .475 

IV.  ENGLISH  INCHES  INTO  MILLIMETERS 476 

V.  U.  S.  STANDARD  AND  SHARP  V-THREADS 477 

VI.  CHORDS  OF  ANGLES  IN  CIRCLES  OF  RADIUS  UNITY  ....  478 

VII.  STANDARD  GAGES  FOR  WIRE  AND  SHEET  METALS     ....  479 

VIII.  SPECIFIC  GRAVITIES  AND  WEIGHTS  OF  SUBSTANCES   ....  480 

IX.  STRENGTH  OF  MATERIALS 481 

X.  FOUR-PLACE  TABLE  OF  LOGARITHMS 482 

XI.  TABLE  OF  NATURAL  AND  LOGARITHMIC  SINES,  COSINES,  TAN- 
GENTS, AND  COTANGENTS  OF  ANGLES  DIFFERING  BY  TEN 

MI.NUTES 484 

INDEX   ,                                                                                                         .  489 


PRACTICAL  MATHEMATICS 


CHAPTER  I 

PRELIMINARY  WORK  AND  REVIEW 

1.  Language  of  mathematics. — Mathematics  has  a  lan- 
guage of  its  own,  with  certain  signs  and  symbols  peculiar  to 
it.  It  is  as  necessary  to  become  familiar  with  these  signs 
and  symbols  and  their  uses,  in  order  to  understand  the  lan- 
guage of  mathematics,  as  it  is  for  the  shorthand  writer  to 
become  familiar  with  the  symbols  used  in  his  work.  Failure 
to  fix  them  in  mind,  and  to  learn  the  definitions  and  technical 
terms  keeps  many  students  from  mastering  the  mathematical 
subjects  they  take  up. 

Some  of  the  best  known  symbols  of  mathematics  are  the  Arabic 
numerals,  1,  2,  3,  4,  5,  6,  7,  8,  9,  0,  the  signs  of  addition,  -)-,  subtraction, 
— ,  multiplication,  X,  division,  -*-,  and  equality,  =,  and  the  letters  of 
the  alphabet.  Other  symbols  will  be  explained  as  used. 

In  the  study  of  mathematics  much  time  should  be  devoted: 

(1)  to  the  expressing  of  verbally  stated  facts  in  mathematical 
language,  that  is,  in  the  signs  and  symbols  of  mathematics; 

(2)  to  the  translating  of  mathematical  expressions  into  common 
language. 

The  signs  and  symbols  of  mathematics  are  used  for  con- 
venience. They  have  gradually  come  into  use  by  general 
agreement.  In  some  cases  the  symbols  are  abbreviations  of 
words,  but  often  have  no  such  relation  to  the  thing  they  stand 
for.  We  cannot  tell  why  they  stand  for  what  they  do  any 
more  than  we  can  tell  why  the  words  for  cat  and  dog  stand  for 
(he  different  animals  they  do.  They  mean  what  they  do  by 
common  agreement  or  by  definition. 

1 


2  PRACTICAL  MATHEMATICS 

2.  How  to  attack  a  problem.— A  problem  in  mathematics 
should  not  be  attacked  as  a  puzzle.     No  guesswork  has  any 
place  in  its  consideration.     The  statement  of  the  problem 
should  be  clear  and  so  leave  but  one  solution  possible.     This  is 
the  business  of  the  author  or  the  one  who  states  the  problem. 
The  following  points  concern  the  student: 

(1)  The  problem  should  be  read  and  analyzed  so  carefully 
that  all  cjnditions  are  well  fixed  in  mind.     If  the  problem  cannot 
be  understood  there  is  no  use  in  trying  to  solve  it.     Of  course, 
if  the  answer  is  given,  a  series  of  guess-operations  may  obtain 
it,  but  the  work  is  worse  than  useless. 

(2)  In  the  solution  there  should  be  no  unnecessary  work. 
Shorten  the  processes  whenever  possible. 

(3)  Always  apply  some  proof  or  check  to  the  work  if  possible. 
A  wrong  answer  is  valueless.     Accuracy  is  of  the  highest  im- 
portance, and  to  no  one  more  than  to  the  practical  man.     If  a 
check  can  be  applied  there  is  no  need  of  an  answer  being  given 
to  the  problem. 

In  this  text  the  answers  follow  most  of  the  exercises.  They  are  given 
for  the  convenience  of  the  student  in  checking  his  work,  and  great  care 
must  be  taken  not  to  misuse  them.  An  answer  should  never  assist  in 
determining  how  to  solve  a  problem.  It  is  best,  then,  not  to  look  at  the 
answer  till  the  problem  is  solved. 

3.  Definitions. — In  order  to  be  exact  in  ideas  and  statements, 
it  is  necessary  to  give  certain  definitions.     It  would  seem, 
however,  that  for  the  practical  man,  technical  terms  should 
be  omitted  so  far  as  possible.     It  is  usually  sufficient  to  make 
the  term  understood,  though  the  definition  may  not  be  a 
good  one  technically.     In  mathematics  more  than  in  almost 
any  other  subject,  each  word  used  has  a  definite  and  fixed 
meaning. 

The  following  definitions  are  inserted  here  to  help  to  recall 
to  mind  some  of  the  terms  used : 

(1)  An  integer,  or  an  integral  number,  is  a  whole  number. 

(2)  A  factor,  or  a  divisor,  of  a  whole  number  is  any  whole 
number  that  will  exactly  divide  it. 

(3)  An  even  number  is  a  number  that  is  exactly  divisible 
by  2. 

Thus,  4,  8.  and  20  arc  even  numbers. 


PRELIMINARY  WORK  AND  REVIEW  3 

(4)  An  odd   number   is   an   integer    that   is   not    exactly 
divisible  by  2. 

Thus,  5,  11,  and  47  are  odd  numbers. 

(5)  A  prime  number  is  a  number  that  has  no  factors  except 
itself  and  1. 

Thus,  1,  2,  7,  11,  and  17  are  prime  numbers. 

(6)  A  composite  number  is  a  number  that  has  other  factors 
than  itself  and  1. 

Thus,  6,  22,  49,  and  100  are  composite  numbers. 

(7)  A  common  factor,  or  divisor,  of  two  or  more  numbers  is 
a  factor  that  will  exactly  divide  each  of  them.     If  this  factor 
is  the  largest  factor  possible  it  is  called  the  greatest  common 
divisor:  abbreviated  to  G.  C.  D. 

Thus,  4  is  a  common  divisor  of  16  and  24  but  8  is  the  G.  C.  D.  of 
16  and  24. 

(8)  A  multiple  of  a  number  is  a  number  that  is  divisible 
by  the  given  number.     If  the  same  number  is  exactly  divisible 
by  two  or  more  numbers  it  is  a  common  multiple  of  them. 
The  least  such  number  is  called  the  least  common  multiple : 
abbreviated  to  L.  C.  M. 

Thus,  36  and  72  are  common  multiples  of  12,  9,  and  4,  but  36  is  the 
L.  C.  M. 

4.  Rules  for  finding  divisor  of  numbers. — It  is  often 
convenient  to  be  able  to  tell  without  performing  the  division, 
whether  or  not  a  given  number  is  divisible  by  another.  The 
following  rules  will  assist  in  this.  Their  proofs  are  simple  but 
are  not  given  here. 

(1)  A  number  is  divisible  by  2  if  its  right-hand  figure  is  0 
or  one  divisible  by  2. 

(2)  A  number  is  divisible  by  3  if  the  sum  of  its  digits  is 
divisible  by  3. 

Thus,  73245  is  divisible  by  3  since  7+3+2+4+5  =21  is  divisible  by  3. 

(3)  A  number  is  divisible  by  4  if  the  number  represented  by 
its  two  last  digits  on  the  right  is  divisible  by  4,  or  if  it  ends  in 
two  zeros. 

Thus,  87656  is  divisible  by  4  since  56  is  divisible  by  4. 


4  PRACTICAL  MATHEMATICS 

(4)  A  number  is  divisible  by  5  if  the  last  figure  on  the  right 
is  0  or  5. 

(5)  An  even  number  the  sum  of  whose  digits  is  divisible  by 
3  is  divisible  by  6. 

(6)  No  convenient  rule  can  be  given  for  7;  the  best  thing  to 
do  is  to  test  by  trial. 

(7)  A  number  is  divisible  by  8  if  the  number  represented  by 
the  last  three  digits  on  the  right  is  divisible  by  8. 

Thus,  987672  is  divisible  by  8  since  672  is  divisible  by  8. 

(8)  A  number  is  divisible  by  9  if  the  sum  of  its  digits  is 
divisible  by  9. 

(9)  A  number  is  divisible  by  11  if  the  difference  between  the 
sum  of  the  odd  digits  and  the  even  digits,  counting  from  the 
right,  is  divisible  by  11. 

Thus,  47679291  is  divisible  by  11  since  (9+9+6+4) -(1+2+ 
7+7)  =11  is  divisible  by  11. 

Note.     This  rule  is  of  little  value  since  the  division  can  be 
tried  about  as  easily  as  the  rule  can  be  applied. 
The  following  facts  are  of  some  value : 

(10)  A  factor  of  a  number  is  a  factor  of  any  of  its  multiples. 

(11)  A  common  factor  of  any  two  numbers  is  a  factor  of  the 
sum  or  the  difference  of  any  two  multiples  of  the  numbers. 

5.  Relative  importance  of  signs  of  operation. — (1)  In  a 
series  of  operations  denoted  by  the  signs  of  addition,  +, 
subtraction,  —  ,  multiplication,  X,  and  division,  -5-,  the  multi- 
plications must  be  performed  first,  the  divisions  next,  and 
lastly  the  additions  and  subtractions. 

(2)  If  several  additions,  or  several  multiplications,  occur 
together  they  may  be  performed  in  any  order. 

(3)  If    several    subtractions,    or    several    divisions,    occur 
together  they  must  be  performed  in  the  order  in  which  they 
come  from  left  to  right. 

The  rules  as  stated  here  are  in  agreement  with  the  best 
usage  in  algebra  and  in  formulas  used  in  practical  work. 

Examples.  (1)  12+3-2+9+7-3  =  26,  by  performing  the 
operations  in  the  order  in  which  they  occur. 

(2)  120-^3X5X2-^2  =  2,    by   first   performing   the   multi- 


PRELIMINARY  WORK  AND  REVIEW  5 

plications  and  then  the  divisions  in  the  order  in  which  they 
occur. 

(3)     12H-3+8X2-6-7-2+7X2X3-9 

=  12-r3+ 16-  6 -=-2+42-  9 

=  4+16-3+42-9  =  50,  by  first  performing  the  multi- 
plications, then  the  divisions,  and  then  the  additions  and 
subtractions. 

EXERCISES  1 

In  the  exercises  1  to  10,  first  perform  the  multiplications,  then  the 
divisions,  and  finally  the  additions  and  subtractions,  each  in  the  order 
in  which  they  occur. 

1.  14  +  16-3  +  10-4-6  =  ?  Ans.  27. 

2.  16-7-8+4X2X3-16X2-7-4  =  ?  Ans.  18. 

3.  15-2X3-15-=-5+4  =  ?  Ans.  10. 

4.  60-25-=-5  +  15-100-r-4x5  =  ?  Ans.  65. 
6.  17X3+27 -=-3 -40X2 -=-5  =  ?  Ans.  44. 

6.  56-7+525-7-5X7X3  +  15-7X8  =  ?  Ans.   13. 

7.  864-=-12-124-=-31+54-=-27  =  ?  Ans.  70. 

8.  4X27 -=-9X4+9X2-3X6-7-9  =  ?  Ans.   19. 

9.  4963-7-7  +  144-7-72-14X9  =  ?  Ans.  585. 

10.  13X9X62+444-4-17X22  =  ?  Ans.  6891. 

11.  Time  yourself  in  doing  the  following  ten  multiplications: 
(1)347X371.  (6)3249X987. 

(2)  547X682.  (7)  4444X888. 

(3)  433X925.  (8)  8764X2233. 

(4)  986X478.  (9)  9898X4257. 

(5)  3587X729.  (10)  9999X8888. 

12.  Check  your  work  in  the  above  ten  multiplications  by  doing  the 
multiplying,  using  the  first  number  in  each  case  as  the  multiplier. 

13.  Time  yourself  in  doing  the  following  divisions.     Check  your  work 
by  finding  the  product  of  the  divisor  and  quotient,  and  comparing  it 
with  the  dividend. 

(1)  395,883 -=-9.  (5)  4,518.976 -=-784. 

(2)  64,362 -=-17.  (6)  783,783^-4147. 

(3)  306,192 -=-48.  (7)   1,312,748 -=-437. 

(4)  87,168 -=-384.  (8)  4,495,491 -=-499. 

14.  Do  the  following  multiplications  and  check  the  work  by  dividing 
the  product  by  the  multiplier  and  comparing  the  result  with  the  multi- 
plicand : 

(1)  843X329.  (3)  4493X345. 

(2)  4327X987.  (4)  8397X9327. 


6  PRACTICAL  MATHEMATICS 

16.  Do  the  following  divisions  and  check  the  work  by  finding  the 
product  of  the  divisor  and  quotient,  then  adding  the  remainder  and 
comparing  the  result  with  the  dividend: 

(1)  43,9624-97.  (4)  9,372,4684-375. 

(2)  842,6374-233.  (5)  4,343,7644-983. 

(3)  467,2344-487.  (6)  3,784,3284-2345. 

16.  The  cost  of  constructing  275  miles  of  railway  was  $4,195,400 
What  was  the  cost  per  mile?  Ans.  $15,256. 

17.  The  circumference  of  a  drive  wheel  of  a  locomotive  is  22  ft.     How 
many  revolutions  will  it  make  in  going  44  miles  if  there  are  5280  ft.  in 
one  mile?  Ans.  10,560. 

18.  If  a  power  plant  consumes  277  tons  of  coal  at  $3.10  per  ton  each 
day,  what  is  the  cost  of  the  coal  to  run  the  plant  one  year  of  365  days? 
Check  the  work. 

19.  A  hog  weighing  78  Ib.  requires  400  Ib.  of  grain  for  each  100  Ib. 
gain  in  weight.     If  the  price  of  56  Ib.  of  grain  increases  in  price  from  42 
cents  to  $1.54  what  should  be  the  increase  in  price  of  hogs  per  100  Ib.? 

Ans.  $8. 

20.  When  56  Ib.  of  corn  cost  56  cents,  hogs  sold  at  $5.60  per  100  Ib. 
live  weight;  and  when  corn  cost  $1.82  for  56  Ib.,  they  sold  at  $15.10  per 
100  Ib.     How  much  more  or  less  does  the  farmer  make  per  hundred  if  it 
takes  400  Ib.  of  corn  for  each  100  Ib.  increase  in  weight  of  hogs? 

Ans.  In  second  case  50  cents  more. 

21.  How  many  tons  of  silage,  2000  Ib.  per  ton,  should  be  stored  for 
20  cows,  the  intention  being  to  feed  each  cow  40  Ib.  a  day  for  5  months, 
then  30  Ib.  a  day  for  2  months,  then  20  Ib.  a  day  for  2  months?     Con- 
sider 30  days  to  a  month.  Ans.  90. 

22.  Give  three  divisors  of  192.     Give  three  multiples  of  72.     Give  a 
common  divisor  of  144  and  192.     Give  the  greatest  common  divisor 
of  these  numbers.     Give  three  common  multiples  of  15,  8  and  20.     Give 
the  least  common  multiple  of  these  numbers.     Is  there  a  greatest  common 
multiple? 

23.  Find  the  prime  factors  of  each  of  the  following:  1188;  148,225; 
89,964;  36,992,000. 

Solution. 
2)1188 

The  work  is  best  carried  out  by  selecting  the  small-  2)594 

est  prime  factors,  leaving  the  larger  till  later.     The  3)297 

prime  factors  of  the  number  are  all  of  the  divisors  used.  3"\99 

The  prime  factors  of  1188  are  2,  2,  3,  3,  3,  11. 

.•>  jo3 

njTi 

i 

24.  Tell  which  of  the  following  numbers  are  exactly  divisible  by  each 
of   2,  3,  4,  5,  6,  7,  and  9:     324;  7644;  3,645,111;  4550:  3645:  49.875: 
23,147,355. 


PRELIMINARY  WORK  AND  REVIEW  7 

6.  Cancellation. — Often  in  solving  problems,  a  fractional 
form  like  the  following  is  obtained: 

64X25X8X12X17 
48X15X32X17X24' 

If  we  do  all  the  multiplications  above  the  line  and  below 
the  line,  and  then  perform  the  division  which  the  line  indicates, 
we  shall  obtain  the  result.  It  is  often  easy  to  avoid  much  of 
this  work,  however,  by  applying  a  principle  of  fractions.  The 
process  which  is  explained  below  is  called  cancellation. 

?      5 


63  ? 

(1)  It  is  seen  that  17  is  found  both  above  and  below  the 
line;  we  draw  a  line  through  each  of  these.     These  numbers 
are  then  said  to  be  cancelled. 

(2)  Now  notice  that  the  numbers  64  and  32  are  divisible 
by  32.     Cancel  64  and  32  and  place  2  above  64  which  is  the 
number  of  times  32  is  contained  in  64. 

(3)  Next,  divide  48  and  8  by  8  and  cancel  them,  writing 
the  quotient  6  below  48. 

(4)  Divide  25  and  15  by  5  and  cancel  them,  writing  the 
quotient  5  above  25  and  3  below  15. 

(5)  In  a  similar  manner  12  and  24  are  cancelled;  also 2 and  2. 
In  this  manner  we  have  replaced  the  given  form  by  the 

5        5 

simpler  one        0  =  j^  •    This  is  the  answer. 
D  Xo      ID 

It  should  be  noted  that  when  no  factor  remains  either  above 
or  below  the  line  after  the  cancellation  is  finished,  we  retain 
one  of  the  unit  factors  which  we  neglected  to  write  when 
cancelling. 

_i.  A 

"9  Ans- 

Remark.  It  should  be  remembered  that  this  method  of 
simplifying  cannot  be  used  when  there  are  additions  or  sub- 
tractions indicated  in  the  problem. 

In  such  a  case  the  operations  above  the  line  must  be  per- 


8  PRACTICAL  MATHEMATICS 

formed  first,  then  those  below,  and  lastly  the  result  above  must 
be  divided  by  the  result  below. 

4+200-6X2      192 


These  processes  may  be  restated  in  the  following: 
RULE.  (1)  Any  factor  above  may  be  divided  into  any  factor 
below  the  line. 

(2)  Any  factor  below  may  be  divided  into  any  factor  above 
the  line. 

(3)  Any  factor  common  to  factors  one  above  and  one  belou 
may  be  divided  into  each. 

(4)  The  answer  is  obtained  by  dividing  the  product  of  the 
numbers  remaining  above  the  line  by  the  product  of  the  numbers 
remaining  below  the  line.     If  no  number  remains  above  or 
below,  use  1. 

EXERCISES  2 

Use  cancellation  to  find  the  results  in  the  following: 

5X8X3X16  20X56X12 

*'      8X15X4  3>  21X10X18 

_    57X119X16  77X100X18X14 

3'  17XI2X19~'  fc  25X11X49X16" 

18X_100XJ3ja2  16X12X7X11 

26X25X9X3    '  *'  24X22X7X18* 

7    90X89X88X87  Ans.         8    1200X515X70X100           Ans. 

1X2X3X4     '    2555,190.  5X35X103                240,000. 

180X132X140X75  Ans.    10    750X4500X5760             . 

15X70X44X36    '  150.            2400X750X50* 

11    Ifj!.  XI  728X999  Ans.    ia    1320X432X660 

96  X  270  X  33    '  290|  |.         4400  X  297  X  288' 


,,'^±45X4         ,„,    ^. 

IB    '6+3X7-4X2  +  167    Ans.  16    256X6  +  125X3-14X76       Ans. 
94-7X9+3X6  4.  17X27+32X40-1618  '         7. 

Analyze  the  following  and  shorten  the  computation  as  much  as 
(lOMsiblc  by  cancellation. 

17.  If  18  men  can  do  a  piece  of  work  in  14  days,  how  many  men  will 
do  the  work  in  21  days? 

Analysis.     If  18  men  can  do  a  piece  of  work  in  14         Operation. 
days,  one  man  can  do  the  work  in  18X14  days.     It         62 
will  take  an  many  men  to  do  the  work  in  21  days  as  21 
is  contained  times  in  18  X  14. 

Hence  it  takes  12  men  to  do  the  work  in  21  days. 


PRELIMINARY   WORK  AND  REVIEW  9 

18.  A  man  worked  16  days  for  30  bushels  of  potatoes  worth  88  cents  a 
bushel.     What  did  he  earn  per  day?  Ans.  $1.65. 

19.  How  many  days  at  $1.50  must  24  men  work  to  pay  for  360  bushels 
of  wheat  worth  $1.20  a  bushel?  Ans.  12. 

20.  How  many  acres  of  potatoes  yielding  150  bushels  to  the  acre  and 
worth  25  cents  a  bushel  will  amount  to  as  much  as  65  acres  of  wheat 
yielding  18  bushels  to  the  acre  and  worth  $1.05  per  bushel?      Ans.  32||. 

21.  If  8  men,  in  15  days  of  10  hours  each,  can  throw  1000  cu.  yd.  of 
earth  into  wheelbarrows,  how  many  men  will  be  required  to  throw  2000 
cu.  yd.  of  earth  into  wheelbarrows  in  20  days  of  8  hours  each? 

Ans.   15. 

Suggestion.  Analyze  the  problem  and  state  in  the  following  form  for 
cancellation: 

8X15X10X2000 
20X8X1000 

22.  A  gardener  sells  75  crates  of  berries,  24  boxes  in  a  crate,  at  8  cents 
a  box,  and  receives  in  return  12  rolls  of  matting,  40  yards  in  a  roll.     Find 
the  price  of  the  matting  per  yard.  Ans.  30  cents. 

23.  A  merchant  bought  15  car-loads  of  apples  of  212  barrels  each, 
3  bushels  in  each  barrel,  at  45  cents  per  bushel.     He  paid  for  them  in 
cloth  at  25  cents  a  yard.     How  many  bales  of  500  yd.  each  did  he  give? 

Ans.  34  and  172  yd.  over. 

24.  How  many  bushels  of  potatoes  at  55  cents  a  bushel  must  be  given 
in  exchange  for  44  sacks  of  corn,  each  containing  2  bushels,  at  30  cents  a 
bushel?  Ans.  48. 

7.  Applying  rules. — The  practical  man  often  has  to  apply 
a  rule  in  solving  a  problem.  This  rule  may;  be  given  to  him 
by  a  fellow  workman,  or  it  may  be  taken  from  a  handbook. 
The  rule  may  be  one,  the  reasonableness  of  which  is  apparent, 
but  often  it  is  not.  Many  rules  are  the  results  of  experience, 
others  of  experiment,  and  still  others  are  mere  "rules  of 
thumb,"  that  is,  they  merely  state  a  combination  of  numbers 
which  gives  the  result  desired. 

In  the  following  problems,  read  the  rule  carefully  before 
applying  it. 

RULE.  To  find  the  number  of  revolutions  of  a  driven  pulley  in  a  given 
time,  multiply  the  diameter  of  the  driving  pulley  by  its  number  of  revo- 
lutions in  the  given  time  and  divide  by  the  diameter  of  the  driven  pulley. 

26.  A  pulley  48  in.  in  diameter  and  making  65  revolutions  per  minute 
(R.  P.  M.)  is  driving  a  pulley  26  in.  in  diameter.  Find  its  number  of 
R.  P.  M.  Ans.  120. 

26.  Find  the  R.  P.  M.  of  a  pulley  8  in.  in  diameter  driven  by  a  28-in. 
pulley,  making  36  R.  P.  M.  Ans.  126. 


10  PRACTICAL  MATHEMATICS 

27.  Find  the  R.  P.  M.  of  a  pulley  44  in.  in  diameter  driven  by  a  32-in. 
pulley,  making  60  II.  P.  M.  Ans.  48. 

RULE.  To  determine  the  width  of  belt  required  to  transmit  a  given 
horse-power  at  a  given  speed  of  the  belt:  For  single  leather  or  4-ply  rubber 
belts,  multiply  the  number  of  horse-power  to  be  transmitted  by  33,000 
and  divide  the  product  by  the  product  of  the  speed  of  the  belt,  in  feet 
per  minute,  multiplied  by  60.  The  quotient  will  be  the  width  of  the  bolt 
in  inches. 

28.  What  is  the  required  width  of  belt  to  transmit  100  horse-power 
with  a  belt  speed  of  3500  ft.  per  minute? 

110 


29.  Find  the  width  of  a  single  leather  belt  to  transmit  75  horse-power, 
with  a  belt  speed  of  3000  ft.  per  minute.  Ans.  13}  in. 

30.  For  heavy  double  leather  or  6-ply  rubber  belts,  use  100  instead 
of  60  in  the  rule.     Find  the  width  of  such  a  belt  to  transmit  135  horse- 
power with  a  belt  speed  of  3600  ft.  per  minute.  Ans,  12|  in. 

RULE.  To  determine  the  horse-power  a  belt  of  given  width  will 
transmit  when  running  at  a  given  speed  :  For  single  leather  or  4-ply  rubbei 
belts,  multiply  width  of  belt  in  inches  by  60  and  the  product  by  speed  ol 
belt  in  feet  per  minute  and  divide  the  product  by  33,000.  The  quotient 
will  be  the  number  of  horse-power  that  the  belt  will  transmit  with  safety. 

31.  How  many  horse-power  will  a  10-in.  single  leather  belt  transmit. 
if  running  at  4000  ft.  per  minute?  Ans.  72^. 

32.  How  many  horse-power  will  a  36-in.  heavy  double  leather  belt 
transmit,  running  at  4500  ft.  per  minute?     (Use  100  instead  of  60  in 
the  rule.)  Ans.  491  nearly. 

The  first  letter  of  a  word  is  often  used  in  mathematics  instead  of  th( 
word  itself.     When  two  or  more  such  letters  are  written  together  with  nc 
sign  between  them  it  is  understood  that  multiplication  is  indicated. 
If  H  stands  for  horse-power, 

P  for  effective  pressure  in  pounds  of  steam  per  square  inch, 
L  for  length  of  piston  stroke  in  feet, 
A  for  area  of  piston  in  square  inches, 
and  N  for  number  of  strokes  per  minute, 

then  the  rule  for  finding  the  horse-power  of  a  steam  engine  may  be  stated 
»n  the  following  abbreviated  form: 

PLAJf. 
"  33000 

33.  Find  //  if  P  =  55  Ib.  per  square  inch,  L  -2  ft.,  A  -  195  sq.  in.,  and 
AT  =80. 

55X2X195X80     _0 
Solutwn.  H  =  -  —         ---  52. 


PRELIMINARY  WORK  AND  REVIEW  11 

34.  Find  H  if  P  =  70,  L  =  2,  A  =  1 65,  and  2V  =  90.  Ans.  63. 

35.  Find  H  if  P  =85,  A  =95,  L  =2,  and  TV  =  190.       Ans.  93  nearly. 

36.  A  railroad  uses  2,240,000  ties  each  year.     If  350  trees  grow  on  one 
acre  and  three  ties  are  cut  from  a  locust  tree  that  is  30  years  old,  how 
many  acres  of  locust  trees  must  be  planted  each  year  to  supply  the  ties? 

Ans.  2133|. 

37.  If  9  men  can  cut  28  cords  of  wood  in  4  days  of  6  hours  each,  how 
many  cords  can  15  men  cut  in  16  days  of  9  hours  each?  Ans.  280. 

38.  A  marble  slab  20  feet  long,  5  feet  wide,  and  4  inches  thick  weighs 
850  pounds.     What  is  the  weight  of  another  slab  of  the  same  marble 
16  feet  long,  4  feet  wide,  and  2  inches  thick?  Ans.  272  pounds. 

39.  If  24  men  in  18  days  of  8  hours  each  can  dig  a  ditch  95  rods  long, 
12  feet  wide,  and  9  feet  deep,  how  many  men  in  24  days  of  12  hours  each 
will  be  required  to  dig  a  ditch  380  rods  long,  9  feet  wide,  and  6  feet  deep? 

Ans.  24. 


CHAPTER  II 

COMMON  FRACTIONS 

DEFINITIONS  AND  GENERAL  PROPERTIES 

8.  The  number  6  when  divided  by  3  gives  a  quotient  of  2. 
This  may  be  written  f  =  2.     If  now  we  attempt  to  divide  6 
by  7,  we  are  unable  to  find  the  quotient  as  above.     The  divi- 
sion may  be  written  $.     This  is  called  a  fraction. 

\  means  that  a  unit  is  divided  into  7  equal  parts.     The 
fraction  4  indicates  that  6  of  the  7  equal  parts  are  taken. 

9.  Definitions. — A  fraction  is  an  indicated  division,  which 
in  a  simple  form  expresses  one  or  more  of  the  equal  parts  into 
which  a  unit  is  divided. 

The  divisor  or  the  number  below  the  line  in  the  fraction  is 
called  the  denominator  of  the  fraction.  The  denominator 
tells  into  how  many  parts  the  unit  is  divided. 

The  dividend,  or  the  number  above  the  line  in  the  fraction, 
is  called  the  numerator  of  the  fraction.  The  numerator  tells 
how  many  of  the  parts,  into  which  the  unit  is  divided,  are 
taken. 

The  numerator  and  the  denominator  are  called  the  terms 
of  the  fraction. 

The  value  of  a  fraction  is  the  number  that  it  represents. 

10.  Mixed  number. — Just  as  we  have  whole  numbers  and 
fractional  numbers,  so  we  have  numbers  made  up  of  whole 
numbers  and  fractions. 

Thus,  we  may  have  2§  which  is  read  2  and  I  and  means  2  + j. 

Definition.  A  mixed  number  is  one  composed  of  a  whole 
number  and  a  fraction. 

11.  Proper  and  improper  fractions. — If  the  fraction  shows 
fewer  parts  taken  than  the  unit  is  divided  into,  its  value  is 
evidently  less  than  1.     If  the  fraction  shows  as  many  or  more 
parts  taken  than  the  unit  is  divided  into,  the  fraction  is  evi- 
dently equal  to  or  greater  than  1. 

12 


COMMON  FRACTIONS  13 

Thus,  2  shows  fewer  parts  taken  than  the  unit  is  divided  into,  and 
is  less  than  1 ;  f  shows  as  many  parts  taken  as  the  unit  is  divided  into  and 
is  equal  to  1 ;  and  \  shows  more  parts  taken  than  the  unit  is  divided  into, 
and  is  greater  than  1.  Then  f  is  a  proper  fraction,  while  f  and  \  are 
improper  fractions. 

Definitions.  A  proper  fraction  is  one  in  which  the  numera- 
tor is  less  than  the  denominator.  An  improper  fraction  is 
one  in  which  the  numerator  is  equal  to  or  greater  than  the 
denominator. 

It  should  be  noted  that  an  indicated  division  is  often  called 
a  fraction,  even  though  the  division  can  be  performed  exactly, 
that  is,  without  a  remainder. 

Thus,  Jg2-,  ff,  44  are  fractions. 

12.  Comparison  of  fractions. — If  two  fractions  have  equal 
numerators  and  equal  denominators  they  are  evidently  equal 
in  value. 

If  two  fractions  have  equal  denominators,  the  one  that  has 
the  larger  numerator  is  the  greater  in  value.  Explain  why. 

Thus,  of  f  and  f ,  f  is  the  larger. 

If  two  fractions  have  equal  numerators,  the  one  that  has  the 
larger  denominator  is  the  smaller  in  value.  Explain  why. 

Thus,  of  |  and  |,  $  is  the  smaller. 

If  two  fractions  have  both  numerators  and  denominators 
unequal,  their  values  cannot  be  compared  so  easily. 

Thus,  the  values  of  f  and  f  can  be  more  easily  compared  when  the 
fractions  are  changed  to  fractions  that  have  the  same  denominator. 
See  Art.  18. 

13.  In  order  to  get  the  right  viewpoint,  it  is  well  for  the 
student   to   note  that  before  he  took  up  fractions  he  had 
learned  to  add,  subtract,  multiply,  and  divide  whole  numbers; 
here  he  has  new  numbers,  fractions,  to  deal  with.     It  is  now 
necessary  to  learn  how  to  perform  the  fundamental  operations 
on  fractions.     They  must  be  combined  not  only  with  other 
fractions  but  with  whole  numbers.     The  main  thing  in  this 
chapter  is  to  do  these  fundamental  operations.     But  to  do 
these  in  all  cases  it  is  necessary  to  be  able  to  change  the  frac- 
tional numbers  in  various  ways,  that  is,  to  reduce  to  lower 
or  higher  terms,  change  fractions  to  common  denominators, 


14  PRACTICAL  MATHEMATICS 

mixed  numbers  to  improper  fractions,  and  improper  fractions 
to  mixed  numbers. 

The  student  studying  alone  must  determine  for  himself  how 
many  exercises  he  needs  to  do  in  order  that  he  may  secure  the 
necessary  accuracy  and  speed. 

14.  Principles.  —  Since  a  fraction  is  an  indicated  division 
the  following  principles  may  be  stated  for  fractions: 

(1)  Multiplying  or  dividing  both  numerator  and  denominator 
by  the  same  number  does  not  change  the  value  of  the  fraction. 

(2)  Multiplying  the  numerator  or  dividing  the  denominator 
by  a  number  multiplies  the  fraction  by  that  number. 

(3)  Dividing  the  numerator  or  multiplying  the  denominator 
by  a  number  divides  the  fraction  by  that  number. 

15.  Reduction  of  a  whole  or  a  mixed  number  to  an  im- 
proper fraction.  —  Example.     Reduce  5  to  6ths. 

Since  1=|,  5  =  5XJ=V.     Ans.  By  principle  (2). 
Example  2.     Reduce  7$  to  5ths. 

Since  1=1  7  =  7X|  =¥• 


The  three  dots,  .*.,  as  used  above  form  a  symbol  meaning 
hence  or  therefore. 

RULE.  To  reduce  a  whole  number  to  a  fraction  of  a  given 
denominator,  first  change  1  to  a  fraction  of  the  given  denomi- 
nator and  then  multiply  the  numerator  by  the  given  whole  number. 
With  a  mixed  number,  reduce  the  whole  number  to  a  fraction  and 
then  add  to  the  numerator  of  this  fraction  the  numerator  of  the 
fractional  part  of  the  mixed  number. 

16.  Reduction  of  an  improper  fraction  to  a  whole  or  mixed 
number.  —  Example  1.     Reduce  -"»4«-  to  a  whole  number. 
J^  =  32-=-4  =  8.    Ans, 

Example  2.     Reduce  -V"  to  a  mixed  number. 
-Y-  =  47-i-9  =  5§.     Ans. 

RULE.  To  reduce  an  improper  fraction  to  a  whole  or  mixed 
number,  perform  the  indicated  division.  The  quotient  is  the 
number  of  units.  If  there  is  no  remainder,  it  reduces  to  a  whole 
number.  If  there  is  a  remainder,  it  reduces  to  a  mixed  number 
of  which  the  quotient  is  the  whole  number  part  and  the  remainder 
the  numerator  of  the  fractional  part. 


COMMON  FRACTIONS  15 

EXERCISES  3 

1.  Reduce  the  following  numbers  to  sixths :  7,  1 1,  40,  17,  19.     To  thirds 
To  tenths. 

2.  Reduce  the  following  mixed  numbers  to  improper  fractions:  2^, 
7?,  9i,  l>4,  17$,  18»,  22H,  46j 

3.  Reduce  the  following  improper  fractions  to  whole  or  mixed  numbers : 

17     J9      2"      32      49      60      71      97      47      t_58      493      9U96     .1.928      9999      376S4 
V>  Vi  -SO  -J->  -?->  -r>  T>  T8J  IS.   Tl  I    17  )    ?fi5  >      a7~j    3J/J  -J4T-- 

17.  Reduction  of  fractions   to  lowest   terms. — Definition. 
A  fraction  is  in  its  lowest  terms  when  the  numerator  and 
denominator  are  prime  to  each  other,  that  is,  when  there  is  no 
integer  that  will  divide  both  of  them. 

Example.     Reduce  -j7^  to  its  lowest  terms. 

75    _  1  5  —5. 
TT>%  —  Tl—  T 

Since  dividing  both  numerator  and  denominator  by  the 
same  number  does  not  change  the  value  of  the  fraction,  both 
terms  may  be  divided  by  5.  Thus  |f  is  obtained.  Both 
terms  of  this  fraction  are  divided  by  3,  and  4  is  obtained. 
Since  5  and  7  are  prime  to  each  other,  the  fraction  is  in  its 
lowest  terms.  Both  terms  could  have  been  divided  by  15  and 
the  reduction  made  in  one  step. 

RULE.  To  reduce  a  fraction  to  its  lowest  terms,  divide  both 
terms  successively  by  their  common  factors,  or  divide  by  the 
greatest  common  divisor  of  the  terms. 

EXERCISES  4 

Reduce  the  following  fractions  to  their  lowest  terms: 

15,3,14,9.  9  7    ,    25,    SI.    24. 

•       13      12     51     27  ••       35'   3S'   28'  ft 

3.    H'Ji-ii-n-  4.   AVri&'l*- 

5.  HI-  Ans.  |-  6.  m-  Ans.  I- 

7.  fli-  Ans.  t-  8.  US-  Ans.  f 

9.  TV5V  Ans.  f-  10.  ffi-  Ans.  H- 

11.  fil?-  Ans.lt-  12.  r9|g-  Arw.  H- 

13.  Jg|3-  Ans.  $J.  14.  AUf.  Ans.  f 

15.  if|8-  Ans.  1TV  16.  mi'  -^ns.  rfflfc- 

17.  TViWo-  ^ns.  flJ.  18.  Um-  Ans.  HI*- 

19.  Reduce  the  following  per  cents  to  fractions  in  their  lowest  terms : 
(The  sign  %  takes  the  place  of  the  denominator  100).  5%,  10%,  40%, 

25%,  35%,  42%,  45%,  30%,  28%,  75%,  80%,  95%,  98%,  14%. 

18.  Reduction  of  several  fractions  to  fractions  having  the 
same    denominator. — Definition.     Fractions    that   have   the 


16  PRACTICAL  MATHEMATICS 

same  denominator  are  called  similar  fractions  or  fractions 
with  a  common  denominator. 

Example  1.  Reduce  $  and  $  to  fractions  which  have  0 
for  a  denominator. 

The  fraction  £  may  be  changed  to  6ths  by  multiplying 
both  its  terms  by  a  number  which  will  make  the  denominator  6. 
This  will  not  change  the  value  of  the  fraction.  This  multiplier 
is  obtained  by  dividing  6  by  2  which  gives  3. 


Likewise       = 


2X3 
1X2 


3X2 

Example  2.     Reduce  I,  f,  and  |  to  72ds. 
Both  terms  of  ^  are  multiplied  by  72  -f-  9  =  8, 
both  terms  of  f  are  multiplied  by  72-7-8  =  9, 
both  terms  of  £  are  multiplied  by  72-7-6=12. 


RULE.  To  reduce  several  fractions  to  fractions  having  a 
common  denominator,  multiply  both  terms  of  each  fraction  by 
a  number  found  by  dividing  the  common  denominator  by  the 
denominator  of  that  fraction. 

19.  Least  common  denominator.  —  In  the  preceding  the 
common  denominator,  72,  was  given.  Usually  the  de- 
nominator is  not  given  but  we  are  asked  to  reduce  the  given 
fractions  to  fractions  having  a  least  common  denominator. 
When  this  is  the  case  we  find  the  least  common  multiple  of  the 
denominators  of  the  given  fractions,  and  this  is  the  least 
common  denominator  (L.  C.  D.)  for  all  the  fractions. 

Example.     Reduce  $,  -fc,  and  ^f  to  fractions  with  a  L.  C.  D. 

The  L.  C.  M.  of  9,  12,  and  24  is  72.  If  we  divide  72  by 
each  of  the  given  denominators  we  get  the  numbers  to  be  used 
as  multipliers. 


Remark.  Usually  the  fractions  dealt  with  have  such 
denominators  that  their  L.  C.  D.  can  be  seen  by  inspection. 
The  student  should  endeavor  to  determine  it  in  this  way 
wherever  possible.  If  it  cannot  be  seen  by  inspection,  a 
good  way  to  find  it  is  as  follows: 


COMMON  FRACTIONS  17 

RULE.  Divide  the  given  denominators  by  a  prime  number  that 
will  divide  two  or  more  of  them,  then  divide  the  remaining  numbers 
and  the  quotients  by  a  prime  number  that  will  divide  two  or  more 
of  them.  Continue  this  as  long  as  possible.  The  L.  C.  D.  is 
the  continued  product  of  all  the  divisors  and  the  quotients  or 
numbers  left. 

Example.     Find  the  L.  C.  D.  of  ^,  ^  i¥o>  and  if- 

Process.     5)30,  45,  135,  25 

3)6,    9,    27,    5 

3)2,    3,      9,    5 

2,     1,      3,    5 

L.  C.  D.  =  5X3X3X2X3X5-1350.     Ans. 

EXERCISES  6 

Change  as  indicated. 

1.  i,  |,  and  I  to  12ths.  2.  |  and  ',  t-r  42ds. 

3.  f ,  f,  and  f  to  24ths.  4.  |,  f ,  and  f  to  63ds. 

5.  2,  I,  and  f  to  42ds.  6.  2,  f,  f,  and  5  to  SOths. 

7.  f,  f,  |,  and  2V  to  lOOths.  8.  f ,  g,  i,  and  |  to  120ths. 

9.  if,  A,  |,  and  f  to  208ths.  Ans.     i§f,  Hf ,  HI,  Hf  • 

10.  |,  T5r,  H,  and  &  to  396ths.  ylns.     HI,  HS,  Ml,  sVr 
Change  the  following  to  fractions  having  a  L.  C.  D. 

11.  -|  and  |-  Ans.  T50,  /„•          12.  |  and  T\-  Ans.       f  f,  4^- 
13.  fV  and  if               ^^s-  io,  IS-          14.  T93  and  T\-          ^ns.  Ml,  Hf  • 
15.  |  and  fr             Ans.  f|f,  //s-          16.  f|  and  if-          Ans.  Iff,  f |f - 

17.  f,  Y,  A,  and  ?V 

18.  ^5,  1,  ii  and  f- 

19.  2A,  4P0,  7/r 

20.  Change  the  following  to  lOOths  and  then  write  as  per  cents:  £, 

1318341          3          7          9J          3          "i          9111317J91          37          917 
5j   4)  t»   Sj    o?   BJ    1C>    10}  TOj    10>  20>  "srOj  50>    20)    »0)    ao>    50>   20,  "25,  ^Et  25,  V5i  tt, 

A,H,f8- 

ADDITION  OF  FRACTIONS 

20.  Example  1.    Add  ^  ^  and  H- 
Just  as  7  apples  +  5  apples  +  11  apples  =  23  apples, 
so  7  twelfths  +  5  twelfths  +11  twelfths  =  23  twelfths. 
The  work  may  be  arranged  as  follows : 

Tz  +  T52  + 1 i  =  f  i  =  ITS  •    Aras. 
Example  2.     Find  the  sum  of  -^  -^5,  -^i 

Here  the  fractions  must  first  be  reduced  to  fractions  having 
a  L.  C.  D.  The  L.  C.  M.  of  12,  15,  and  30  is  60. 


18  I'KACTICAL  MATHEMATICS 


W-W-     An*' 
Example  3.     Find  the  sum  of  3J,  5$,  2^,  7J- 
The  whole  numbers  and  the  fractions  may  he  added  sepa- 

rately, and  then  these  sums  united.     The  work  may  t>e  written 

as  here. 


A    more    convenient    way    of  3*    = 

writing  the  mixed   numbers  for  5f   = 

adding,  is  to  write  them  under  2^  = 

each  other,  and  add,  similar  to  7J   = 
the    method    of    adding    whole 


numbers.  Ans. 

RULE.  To  add  fractions  that  have  a  L.  C.  D.,  add  the  numera- 
tors of  the  fractions  and  place  the  sum  over  the  L.  C.  D.  If 
this  gives  an  improper  fraction,  it  should  be  reduced  to  a  whole 
or  mixed  number.  If  the  fractions  do  not  have  a  L.C.D.,  first 
reduce  them  to  fractions  wiih  a  L.  C.  D.  To  add  mixed  num- 
bers, add  the  whole  numbers  and  fractions  separately  and  then 
unite  the  sums. 

EXERCISES  6 

Add  the  following  and  express  the  sum  in  the  simplest  form. 

i.  i+i+j+i-  2.  t  +I+J+Y. 

3.  &+&+H+H-  4.  l+i+A- 

6.  7  +  ?+4i+7J-  6.  9J+3J+6. 

7.  41J+40J+3.  8.  9*+7i+8j- 
9-  A+A  +  I  +  I-  Ans.  1JH- 

10.  f  +  V+l+V  Ans.  Si- 

11. iji+fl+ifi.  Ans.  60. 

12.  V  +  V+A-  Ans.  5W- 

13.  214J+517/J  +  145&-  Ans.  876f|» 

14.  3!  +  17i  +  28A+3,V  Ans.  53A- 

15.  S  +  S  +  l  +  i  +  i+^+A-  Ans.  3|- 

16.  2|+7|+HA+14!+17H-  Ans.  54\. 

17.  871A+614J|+81f  Ans.  1067|j- 

18.  145J+36  +  H  +  194  +  H-  Ans.  376Hf- 

19.  126i  +  35  +  15J+5SJ+9rV  Ana.  245f- 

20.  16|  +  14J  +  17i  +  19j+27A-  Ans.  95j- 

21.  16^  +  191+24^+29/0+14.  Ans.  1035- 

22.  A  merchant  sold  to  different  customers  5  1  yards  of  cloth,  7J  yards. 
15J  yards,  9}  yards,  and  3|  yards.  Find  the  total  number  of  yards  sold. 

Ans.  41JJ- 


COMMON  FRACTIONS  19 

23.  A  farmer  has  10£  acres  in  one  field,  8|  acres  in  another,  and  30j 
acres  in  a  third.     How  many  acres  in  the  three  fields.          Ans.  495- 

24.  In  five  days  a  steamer  sails  the  following  distances:  384f  miles, 
372 1  miles,  356  5  miles,  392  J  miles,  and  345 1  miles.     How  far  did  it  sail 
in  the  five  days?  Ans.  1852J  miles. 

SUBTRACTION  OF  FRACTIONS 

21.  Example  1.     Subtract  T4r  from  i9i- 

Since  like  numbers  can  be  subtracted  we  can  subtract 
4  elevenths  fiom  9  elevenths  and  have  the  remainder  5 
elevenths.  This  may  be  written  -i9r~~T4r='iV  Ans. 

Example  2.     Subtract  T7T  from  f- 

Here  the  fractions  must  first  be  reduced  to  fractions  having 
the  same  denominator.  It  may  be  written 

2  7    —  .22  21—  JL       A  <n  t 

3  TT~ff      tt~~T8'    A-ns- 

Example  3.     7f  -  3|  =  what?  Solution. 

In   this   case  the  fractional  part  of  the  71  =  710 

subtrahend     is     less     than    that    of    the  03—09° 

minuend.     The   fractional  parts   of  mixed  — - 

numbers  are  reduced   to  fractions  having 
the  L.  C.  D.,  the  fractional  parts  subtracted,  and  then  the 
whole  numbers. 

Example  4.     7\  —  3f  =  what? 

In    this   case  the  fractional   part   of 
the  subtrahend  is  greater  than  the  frac-  Solution. 

tional  part  of  the  minuend.  The  frac-  7^=7f  =6f 
tions  are  changed  to  fractions  having  3f  =  3f  =3|- 
the  L.  C.  D.  as  before.  It  is  then  Ans.  3f 

noticed  that  the  fraction  |  in  the  sub- 
trahend  is  larger   than  f-  in  the  minuend  and  so  cannot  be 
subtracted  from  it.     To  overcome  this  difficulty  we  take  1 
from  the  7  and  change  it  to  sixths.     This  gives  6f  instead 
of  7f .     The  subtraction  is  then  made  as  before. 

RULE.  To  find  the  difference  between  two  fractions  having 
a  common  denominator,  find  the  difference  of  the  numerators 
and  write  it  over  the  common  denominator.  If  the  fractions  do 
not  have  a  L.  C.  D.  reduce  them  to  such  before  subtracting.  If 
the  numbers  are  mixed  numbers,  subtract  the  fractional  parts 
and  then  the  whole  numbers. 


20  PRACTICAL  MATHEMATICS 

EXERCISES  7 

Subtract  the  following  and  give  the  results  in  their  simplest  forms. 

1.  1-i-  2-  l-l                    3.  A-i- 

4-  ft-A  6-  J-A-                    6.  2J-|. 

7.  8}-f  8.  7-4S-                    9.  9i-l$- 

10.  ff-rir  ^w-  2|f-  11.  M-A'          An*.  TH 

12.  U-fr  Ans.  HI-  13.  4|-1/0.          Ana.  2^ 

14.  8i-2|-  Ans.  6/0-  15.  9f-3$-           ^rw.  5Jf- 

16.  463J  17.  346|  18.  461? 

146&  146J                            145| 

19.  469ft  20.  192ft  21.  229  J 

21  &  142A                          163g 

22.  230|  23.  117§  24.  403ft 

103f  96|                           2311 

Simplify  the  following,  that  is,  do  the  operations  indicated: 

25.  12|  +28|-15|-    Ans.  25|g-    28.  4?-?-|+6f  Ans. 

26.  5f+2i-3l-         Ans.  4J-        29.  14+6i-9f-  Ans.  10$ 

27.  4J-21+1J-          Ans.  3gg-      30.  |  +  13-(6J-|)+}-    Ans.  8JJ- 

The  parentheses  indicate  that  the  enclosed  operations  must  be  per- 
formed first.  Thus,  in  the  above,  f  must  be  subtracted  from  6J  before 
they  are  subtracted  from  13. 


31.  ?+17+ft-(6  +  9j)-  Ans.  21- 

32.  4|+3f+6i-(H  +  H>  Ans.  10JJ- 

33.  7f+6J-2f+?+2ft-  Ans.  14\\- 

34.  3i+41  +  l?-(?+22»T)-  Ans.  6^- 

35.  7|+2!-3f+(lf+lJ)-  Ans.  9if- 

Do  as  many  of  the  following  as  you  can  without  a  pencil. 

36.  A  boy  had  $J  and  spent  $i;  how  much  money  did  he  have  le^t? 

37.  A  man  bought  2  J  tons  of  coal  and  had  1}  tons  delivered  ;  how  much 
was  left  to  be  delivered? 

38.  A  man  had  5f  acres  of  land  and  sold  3$  acres;  how  many  acres 
did  he  have  left? 

39.  A  man  weighed  159}  Ib.  on  Monday  and  154$  Ib.  on  the  following 
Saturday;  how  many  pounds  did  he  lose? 

40.  A  man  had  $7$  and  paid  a  debt  of  $3};  how  much  did  he  have  left? 

41.  A  man  sold  J  of  his  farm  at  one  time  and  }  of  it  at  another;  what 
part  of  his  farm  did  he  sell?     What  part  did  ho  have  left? 

42.  A  coal  dealer  had  10  tons  of  coal.     He  sold  3i  tons  to  one  customer, 
2}  tons  to  another  and  the  remainder  to  a  third  customer.     How  much 
did  he  sell  to  the  third  customer? 


COMMON  FRACTIONS  21 

43.  A  tank  full  of  water  has  two  pipes  opening  from  it,  one  will  empty 
5  of  the  water  in  the  tank  in  one  hour  and  the  other  £  of  it;  what  part 
will  both  pipes  empty  in  one  hour?     What  part  remains  in  the  tank? 

44.  Find  the  distance  around  the  figure  with  di-              g, 
mensions  as  given.  __ — v      ,, 

45.  One  coal  wagon  drew  6i7o  and  8|  tons  of  coal      \  \*" 
on  two  successive  days;  another  wagon  drew  7-fg  and  $*! \  / 
9f  tons  on  the  same  days.     How  much  more  did  the        \                /^~ 
latter  draw  than  the  former?                                                    \______-v 

22.  Resultants. — The    combined    effect   of          FIQ*  1 
several  forces  is  called  the  resultant.     Thus, 

a  pull  of  100  Ib.  toward  the  east  and  at  the  same  time  a  pull 
of  75  Ib.  toward  the  west  gives  a  resultant  pull  25  Ib.  toward 
the  east. 

The  resultant  of  pulls  of  150  Ib.  toward  the  east,  85  Ib. 
toward  the  west,  and  75  Ib.  toward  the  west  is  a  pull  of  10 
Ib.  toward  the  west. 

46.  Using  E  for  east,  W  for  west,  N  for  north,  and  S  for  south;  find 
the  resultants  of  the  following: 

(1)  48£  Ib.  W,  92  f  Ib.  E,  76|  Ib.  E,  and  9H  Ib.  W.  (2)  125&  Ib.  N, 
751  Ib.  N,  47f  Ib.  S,  and  156^  Ib.  S. 

Ans.   (1)  29i  Ib.  E;  (2)  3  Ib.  S. 

MULTIPLICATION  OF  FRACTIONS 

23.  Multiplication    of   fraction  and  integer. — -Example  1. 
Multiply  |  by  4. 

To  multiply  f  by  4  is  to  find  a  fraction  that  is  4  times 
as  large  as  f .  By  Art.  14,  multiplying  the  numerator  of  a  frac- 
tion multiplies  the  value  of  the  fraction. 

3X4     12 
.'.  |X4  =  — —  =— =2g-  Ans. 

Example  2.     Multiply  8  by  f  • 

Since  in  finding  the  product  of  two  numbers  either  may  be 
used  for  the  multiplier  without  changing  the  product, 

Example  3.     Multiply  T3^  by  7. 

Here  we  may  use  the  principle  that  dividing  the  denomina- 


22  PRACTICAL  MATHEMATICS 

tor  multiplies  the  value  of  the  fraction,  or  the  operation  may 
be  thought  of  as  one  in  cancellation. 

---  An*. 


OrT\X7  =  ~p  =  $  =  U.  Am. 

Here  the  7  and  14  are  cancelled. 

RULE.  To  multiply  a  fraction  by  an  integer  or  an  integer  by 
a  fraction,  multiply  the  numerator  or  divide  the  denominator 
of  the  fraction  by  the  integer. 

Remark.  When  a  whole  number  is  multiplied  by  a  whole 
number  the  product  is  larger  than  the  multiplicand;  but 
whenever  the  multiplier  is  a  proper  fraction  the  product  is 
smaller  than  the  multiplicand.  Here  we  cannot  think  of 
multiplication  as  a  shortened  addition. 

We  often  write  |  of  6  for  |X6. 

The  meaning  is  the  same  in  each  case. 

24.  Multiplication  of  a  fraction  by  a  fraction.— 

Example  1.     Multiply  f  by  4' 

f  by  f  is  the  same  as  4  of  $,  but  4  of  f  is  5  times  j  of  f  and  | 
of  f  has  a  value  |  as  large  as  f  . 

By  Art.  14,  the  value  of  a  fraction  is  divided  when  the  denomi- 
nator is  multiplied. 


2X5 
And  4  of  f  =  5  times  ^  =  —^- 

These  steps  may  be  combined  as  follows: 

4-0          An 


Example  2.     Multiply  }|  by  f  • 

_l  . 

--     Ans. 


Cancellation  should  be  used  when  it  will  shorten  the  work. 
Example  3.     Multiply  f  by  ^  by  $?• 

2 


COMMON  FRACTIONS  23 

RULE.  To  multiply  a  fraction  by  a  fraction,  multiply  the 
numerators  together  for  the  numerator  of  the  product,  and  the 
denominators  together  for  the  denominator  of  the  product.  Cancel 
when  convenient. 

Remark.  A  form  like  f  of  f  of  f  is  often  called  a  com- 
pound fraction. 

25.  Multiplication  of  mixed  numbers  and  integers.  — 

Example  1.     Multiply  7|  by  6. 


5  5 

Example  2.     Multiply  8|  by  3| 


?-     Ans. 

8|  by  3|- 
5 


O  p  O 

RULE.  To  multiply  two  numbers,  one  or  both  of  which  are 
mixed  numbers,  reduce  the  mixed  numbers  to  improper  frac- 
tions and  multiply  as  with  fractions. 

Remark.  The  work  may  often  be  simplified  by  using  the 
following  methods : 

Example  3.     Multiply  47  by  16f  • 

Process. 

47  Explanation.     Multiply   47   by   4   and    divide 

16|      by  5,  which  is  the  same  as  multiplying  47  by  |; 

5)188        this  gives  37f  •     Then  multiply  47  by  16  in  the 

37§       ordinary    way   for    multiplying    whole   numbers. 

Add  these  three  partial  products  and  the  entire 

__        product  is  789|  • 
789f 

Example  4.     Multiply  25 f  by  6^- 

Process.         Explanation,     f  X  i  =  & ',  25  X  \  =  8-| ;  |  X  6  =  2| ; 
25f      25X6=150. 

Al 

The  entire  product  equals  the  sum  of  these 


QI      partial  products. 


2| 
150 

160U 

If  several  fractions  and  mixed  numbers  are  to  be  multiplied 
together  it  is  usually  best  to  reduce  all  to  fractions  for  then  the 
work  may  be  shortened  by  cancellation. 


24  PRACTICAL  MATHEMATICS 

Example  5.     Find  the  product  of  J  X3§X9X412, 

23 


'    "     __  •*•"•*•  __  |   A    ij  i 

11 


EXERCISES  8 
Find  the  product  of  each  of  the  following;  without  pencil  when  possible. 


1.  1X4.  10.  25XA-  19. 

2.  JX4.  11.  5XA-  20.  lof  20. 

3.  $X2.  12.  15  XH  21.  I  of  30. 

4.  $X5.  13.  27Xli  22.  f  of  63. 
6.  AX8.                     U.  45X2J-                      23.  A  of  120. 

6.  AX5-  16.  55X2J-  24.  f  of  99. 

7.  7X|-  16.  iof  I  25.  A  of  22. 

8.  8X|-  17.  SXJ-  26.  6iX8. 

9.  9Xf  18.  AX  A-  27.  4AX6. 

28.  What  is  3  times  4  bushels?     3  times  4-fifths?     3  times  }?     3  X  t*i  ? 

29.  What  is  $  of  9  quarts?     $  of  9-tenths?     $  of  A?     J  Xrj? 

30.  A  can  is  3  full  of  milk.     If  I  of  this  is  drawn  off,  what  part  of  the 
whole  can  is  drawn  off?     What  part  remains  in  the  can? 

31.  It  took  a  boy  living  in  the  country  50  minutes  to  walk  to  school. 
He  could  drive  with  a  horse  in  f  of  this  time.     How  long  did  it  take 
him  to  drive? 

32.  On  one  field  a  farmer  harvested  230  bushels  of  wheat  and  on  a 
second  f  as  much.     How  many  bushels  were  harvested  on  the  second 
field? 

33.  One-third  of  the  water  in  a  tank  will  flow  from  a  certain  opening 
in  1  hour.     If  the  tank  holds  60  barrels  how  many  barrels  will  flow  out 
in  2  hours? 

34.  If  two  pipes  open  from  a  tank,  one  of  which  can  empty  J  of  the 
tank  in  1  hour,  and  the  other  J  of  it  in  1  hour,  what  part  of  the  tank  will 
both  empty  in  1  hour?     If  the  tank  holds  60  barrels,  how  many  barrels 
will  flow  out  in  2  hours  if  both  pipes  are  open?  Ans.  Ai  54. 

35.  The  circumference  of  a  circle  is  about  3f  times  the  diameter. 
Find  the  circumference  of  a  circle  if  the  diameter  is  7  ft.,  21  ft.,  6  ft. 

36.  The  diagonal  of  a  square  is  very  nearly  1  ,5j  the  length  of  one  side. 
Find  the  diagonal  when  one  side  is  12  in.,  6  in.,  84  ft. 

37.  If  you  have  a  vacation  of  100  days,  and  spend  i  in  the  country, 
A  camping,  and  the  remainder  in  the  city,  how  many  days  do  you  spend 
in  each  place? 

38.  A  boy  has  $2.50.     He  spends  I  of  it  for  a  fishing  rod,  A  of  it  for  a 
reel,  and  the  remainder  for  a  line.     How  much  did  he  spend  for  each? 


COMMON  FRACTIONS  25 

39.  A7sXl6.  Ans.  IfJ.  40.  7V,X96.  Ans.  6T83. 

41.  fiiX96.  Ans.  62£.  42.  28T52X14.         Ans.  397«. 

43.  816§X17.        Ans.  13883£.       44.  956*  X29.         Ans.  277471. 
45.  12^X62§.        Ans.  781i          46.  12fX3f.          Ans.  47||. 
47.  13iX9f.          Ans.  131|.          48.  23|Xl8f.         Ans.  441|. 
49.  14|X10|.         Ans.  153ff.         50.  212fX7i         Ans.  1595. 

51.  Multiply  1|  by  1§,  2J  by  2J,  3|  by  3J,  10J  by  10J. 

52.  Can  you  make  a  rule  for  finding  the  product  of  two  factors  that 
are  the  same  and  end  in  5?     See  Art.  43  (5). 

The  product  of  two  factors  that  are  exactly  alike  is  called  the  square  of 
one  of  them.  Thus,  the  square  of  4 J  is  4|  X4|  =20j- 

Find  the  square  of  each  of  the  following  by  your  rule:  7%,  9J,  11 5, 
16J,  m,  20|,  100|. 

53.  fXfXf=what?  Ans.  TV 

54.  fXfXlfXl=what?  Ans.  f. 

55.  AX2jX7f  X29gX3|=what?  Ans.  7£f- 

56.  If  hogs  are  worth  9f  cents  a  pound,  what  is  a  hog  weighing  325  lb. 
worth?  Ans.  $31.68f. 

57.  In  Chicago  in  1912,  carpenters  received  65  cents  per  hour.     How 
much  was  this  for  an  8-hour  day?     For  one  week  of  5^  days? 

Ans.  $5.20;  $28.60. 

58.  In  the  same  city,  a  bricklayer  received  72 1  cents  an  hour.     How 
much  could  a  man  earn  in  a  year  if  he  worked  225  days  of  8  hours  each  ? 

Ans.  $1305. 

59.  In  1908,  a  stonecutter  in  New  York  received  56/0  cents  an  hour. 
If  a  man  worked  50  weeks  a  year  and  5j  days  per  week,  8^  hours  a  day, 
how  much  would  he  earn  a  year? 

Solution.     56/o  X8^  X5|  X50  =  HJP  X1/-  XV-  X¥  =$1312.50f .     Ans. 

60.  A  man  earns  62  J  cents  an  hour  and  his  two  sons  each  22 f  cents  an 
hour.     How  much  do  the  three  earn  per  week  of  5J  days  of  8?  hours  each? 

Ans.  $50.49. 

61.  A  gang  of  men  mix  and  place  an  average  of  43J|  cu.  yd.  of  con- 
crete per  hour.     How  many  cubic  yards  do  they  place  in  a  day   of  8J 
hours?  Ans.  381. 

62.  An  alloy,  used  for  bearings  in  machinery,  is  if  copper,  ^  tin,  and 
^V  zinc.     How  many  pounds  of  each  in  346  lb.  of  the  alloy? 

.4ns.  286^;  47|J;  llfi 

63.  An  alloy,  called  "anti-friction  metal,"  is  ylou  copper,   \\%  tin, 
and  j3o  antimony.     Find  the  weight  of  each  metal  in  a  mass  of  the  alloy 
weighing  1250  lb.  Ans.  46 £  lb.;  1110  lb.;  93 f  lb. 

64.  Find  the  cost  of  27f  sq.  ft.  of  plate  glass  at  66|  cents  per  square 
foot.  Ans.  $18.40. 

65.  A  pumping  engine  in  Chicago  pumps  on  an  average  of  17, 361  \ 
gallons  per  minute,  how  many  gallons  is  this  in  24  hours? 

Ans.  25,000,000. 

66.  An  ice-plant  has  an  output  of  45  tons  daily.     What  is  the  value  of 
this  output  for  a  year  of  320  days  at  $3g  per  ton?  Ans.  $51,840. 


26 

67.  Nickel  ateel  will  stand  a  pull  of  90,000  Ib.  per  square  inch  of  cross 
section.     What  pull  will  a  bar  of  l|f  sq.  in.  cross  section  stand? 

ATM.  174,375  Ib. 

68.  The  average  yearly  fire  loss  in  the  United  States  from  1897  to 
1906  was  $2i7(j  per  capita.     If  the  population  averaged  75,000,000,  what 
was  the  average  loss  per  year?  Ana.  $202,600,000. 

69.  In  the  European  countries  for  the  same  period  as  in  the  previous 
exercise,  the  average  fire  loss  per  capita  was  $J.     What  would  have  been 
saved  in  the  United  States  if  the  fire  loss  had  been  the  same  as  in  the 
European  countries?  Ans.  $177,500,000. 

70.  The  circumference  of  a  circle  is  very  nearly  f?f  times  the  diameter. 
What  is  the  circumference  of  a  circle  that  is  24/0  in.  in  diameter? 

Ans.  77ii|  in.  nearly. 

71.  If  the  diagonal  of  a  square  is  very  nearly  1^  the  length  of  one  side, 
find  the  diagonal  in  feet  of  a  square  \\  miles  on  a  side.     A  mile  is 
5280  ft.  Ans.  50CO  ft. 

72.  Remembering  that  6  per  cent  means  rfof,  find  the  value  of  the 
following  : 

(a)  6%  of  $7.25.  (/)   45%  of  325  acres. 

(6)  9%  of  $820.  (0)  75%  of  $3276. 

(c)  12%  of  $75.20.  (h)  95%  of  396  miles. 

(d)  20%  of  476  bushels  (i)'  82%  of  7684  bushels. 

(e)  30%  of  9227  bushels.  0')   65%  of  4762. 
Ans.     (o)  $0.43  J;  (e)  2768  &  bu.;  (g)  $2457;  (j)  3095^. 

DIVISION  OF  FRACTIONS 

26.  Division    of   a   fraction   by   an   integer.  —  Example    1. 
Divide  4  by  4. 

(1)  Since  to  divide  by  4  is  to  find  one  of  the  4  equal  parts 
and  to  get  \  of  a  number  is  to  find  one  of  the  4  equal  parts, 
we  have 

f-s-4  =  i  off  =&•     Ans. 

(2)  Or,  using  the  principle  that  multiplying  the  denominator 
of  a  fraction  divides  the  value  of  the  fraction,  we  have 


(3)  In  division  of  fractions,  we  can  often  divide  the  numera- 
tor and  thus  divide  the  fraction. 
Example  2.     Divide  Yi4  by  25. 

925  925^-25    37 

TT         -IT  =TT=3*' 

This  may  be  written  W^-25  =  WX-sVHi  =3^-     Ans. 


COMMON  FRACTIONS  27 

RULE.  To  divide  a  fraction  by  an  integer,  divide  the  numerator, 
or  multiply  the  denominator,  of  the  fraction  by  the  integer;  or 
multiply  the  fraction  by  1  over  the  integer. 

27.  Division  by  a  fraction.  —  Example  1.     Divide  6  by  f 

(1)  If  we  reduce  6  to  thirds,  we  may  divide  the  numerators, 
since  then  the  numbers  will  both  be  thirds,  and  so  be  like 
numbers. 

6  -5-f=  -^-5-1  =  18  thirds  ^-2  thirds  =9.     Ans. 

(2)  Since  there  are  3  times  ^  in  1,  and  |  as  many  times 
|»  there  are  f  times  3,  or  f  times  f  in  1.     Now  -f  is  f  in- 
verted.    Hence  we  can  find  how  many  times  the  fraction 
|   is   contained   in   1   by  inverting  the  fraction,     f  will  be 
contained  6  times  as  many  times  in  6  as  it  is  contained  in  1. 

.-.  6-hf  =  6Xf  =  9.     Ans. 

Definition.  The  reciprocal  of  a  number  is  1  divided  by  that 
number.  Thus,  f  is  the  reciprocal  of  f.  f  is  the  recipro- 
cal of  f  or  4. 

Example  2.     Divide  c§  by  £|. 


49  _i_  1  4  —  49V39  —  21—  O   1 
65    •   39"  ~  (55  A  1¥~  "17  —  ^1  IT 

Example  3.     Divide  4|  by  3|. 

First  reducing  each  to  improper  fractions,  we  have 
4*-3-3$=¥-¥  =  Y-XA  =  -H  =  ltt.     Ans. 

RULE.  To  divide  a  whole  number  or  a  fraction  by  a  fraction, 
invert  the  divisor  and  multiply  by  the  dividend.  If  either  or 
both  dividend  and  divisor  are  mixed  numbers,  first  change  to 
improper  fractions. 

28.  Special  methods  in  division.  —  The  work  of  division  may 
often  be  simplified  by  one  of  the  following  methods: 

Example  1.     Divide  56  f  by  5. 

Process.  Explanation. 

5)56f  56-5-5  =  11  with  a  remainder  of  1. 

1H     Ans.         ]f-=-5  =  f-=-5  =  ^- 

Example  2.     Divide  75  by  3|  • 

Process.  Explanation.     Since  multiplying  both 

dividend    and    divisor    by    the    same 

3  1)  75  number  does  not  change  the  quotient, 

11)225  we  can  multiply  both  by  the  denomi- 

20/1      Ans.          nator    in    the  divisor,   then   divide  as 
before. 


28  PRACTICAL  MATHEMATICS 

Example  3.     Divide  125§  by  2f  • 

Process.  Explanation.  The  same  as  in  the 

2f)125§  preceding,  multiply  both  by  the  denomi- 

lll**P_?J_  nator  in  the  divisor.  Then  divide  as  in 

45|f  Am.  example  6. 

EXERCISES  9 

Divide  the  following,  using  the  pencil  only  when  necessary: 


1.    f-5- 

4. 

6. 

ljV  -5-H. 

11.  * 

j-j. 

2.  \\- 

-3. 

7. 

_W"X12. 

12.  f- 

LA- 

3.  Y- 

-3. 

8. 

5-^-i' 

13.  l- 

5-A' 

4.  -fi  - 

-4. 

9. 

17+f- 

14.  1*3 

-5-1. 

6.  ?1- 

-7. 

10. 

16-5-f. 

15.  Y 

•«-¥ 

16. 

32  J-  4. 

21. 

961-5-8. 

17. 

326  f-5-  2. 

22. 

122|-3. 

18. 

764?  -4 

23. 

27|^-5. 

19. 

211-5-2. 

24. 

86f  -=-3. 

20. 

28!  -r3. 

25. 

192|-i-5. 

26.  If  the  denominator  of  a  fraction  is  multiplied  by  3,  how  is  the  unit 
of  the  fraction  changed?     How  changed  if  multiplied  by  6?     By  5? 
Illustrate  with  the  fraction  §. 

27.  If  \  in.  on  a  map  represents  1  mile,  how  many  miles  are  represented 
by  6  in.  on  the  map? 

28.  In  the  drawing  of  a  house,  \  in.  in  the  picture  represents  1  ft.  in 
the  actual  house.     Find  the  dimensions  of  the  rooms  that  measure  as 
follows:  2J  in.  by  2|  in.,  1}  by  1},  1&  by  1^,  ft  by  ft. 

In  the  following,  x  is  used  for  the  number  that  is  to  be  found : 

29.  §-{-6  =  x.  32.  x-s-J  =  7.  35.  j  -s-4  =  -• 

x 

30.  3-5-z  =  6.  33.  Jp-ns  =  ?.  36.  y-r-9  =  — . 

\s  X 

31.  |*.r  =  5.  34.  V-l  =  g-  37.  ?-hx  =  ?- 

38.  If  a  man  can  do  a  piece  of  work  in  3  hours,  what  part  of  it  can  he 
do  in  1  hour?     In  2  hours? 

39.  If  a  man  can  do  i  of  a  piece  of  work  in  1  hour,  in  how  many  hours 
can  he  do  all  the  work?     §  of  the  work? 

40.  If  a  man  can  do  f  of  a  piece  of  work  in  2  hours,  in  how  many 
hours  can  he  do  all  the  work?    f  of  the  work?     J  of  the  work?    A  of 
the  work? 

41.  If  a  boy  can  run  §  of  a  mile  in  6  minutes,  how  many  minutes  will 
it  take  him  to  run  a  mile?     J  a  mile?     f  of  a  mile? 


COMMON  FRACTIONS  29 

42  If  John  can  do  ^  of  a  piece  of  work  in  1  hour  and  Henry  can  do  £ 
of  it  in  1  hour,  what  part  of  the  work  can  they  both  do  in  1  hour?  How 
many  hours  will  it  take  them  to  do  the  whole  work  if  they  work  together? 

Ans.  f  ;  1£. 

43.  One  boy  can  hoe  a  patch  of  potatoes  in  3  hours  and  another  boy 
can  hoe  the  same  patch  in  4  hours.     In  how  many  hours  can  they  hoe 
the  potatoes  if  they  work  together? 

Solution.  The  first  boy  can  hoe  |  of  the  patch  in  one  hour,  and  the 
second  boy  J  of  the  patch  in  one  hour.  Together  they  can  hoe  §  +  }  =TV 
of  the  patch  in  one  hour.  They  can  hoe  the  entire  patch  in  l-=-i7j  = 
lXV=¥  =  lf  or  If  hours. 

44.  A  water  tank  that  holds  60  barrels  has  two  pipes  opening  from  it. 
One  of  these  can  empty  the  tank  in  4  hours  when  running  alone,  and  the 
other  pipe  can  empty  the  tank  in  12  hours  when  running  alone.     If  both 
of  the  pipes  are  running  at  the  same  time,  how  long  will  it  take  them  to 
empty  the  tank?  Ans.  3  hours. 

45.  If  one  pipe  can  empty  a  tank  in  4  hours,  and  another  pipe  can 
empty  it  in  12  hours,  in  how  many  hours  will  both  pipes  empty  the  tank 
when  running  together?  Ans.  3. 

46.  If  a  boy  earns  SJ  a  day,  in  how  many  days  will  he  earn  $9. 

Ans.  12. 

47.  If  a  man  earns  $|  in  1  hour,  in  how  many  hours  will  he  earn  $15? 

Ans.  10. 
48. 
49. 
60. 

51.  100  -4| 
52. 

53.  31|  -I 
54. 

In  the  following  five  exercises,  reduce  all  to  fractions,  take  reciprocals 
of  each  divisor  and  cancel. 
62.  2^X31  Xr4rX2r^-20f. 

Process. 

2 
Xa+20f-xxxx-.     Ans. 


Ans.  ^V. 

55.  3}i-^7|f. 

A  -MO      295 

/ins.  5^7. 

Ans.  |f. 

56.  T&r-TriT. 

Ans.  2f4f. 

Ans.  !TvV 

57.  lOH-5-lrf*. 

Ans.  10. 

Ans.  20  1$. 

58.  51-^3- 

Ans.  24|. 

Ans.  T95. 

59.  7^3MJ. 

Ans.  Iff. 

Ans.  18. 

60.  104T3TH-8&. 

Ans.  11U. 

Ana      1253 
/I/IS.     l^j^. 

61.  1151-5-20$. 

Ana.  5f. 

63.  SiXQg-T-Sif-s-Qf-s-S&^-riy.  Ans. 

64.  if  X29f  Xl3f-^15f-r-2|.  Ans. 

65.  &X  13*  X*|X9i  -5-1!  •*-(!&  XfX26i)-H.  Ans.  2ff. 

66.  8f  XHX&XHXHX71X  11-5-6$  -Hf  *12i-5-£.  Ans.  1*. 

67.  Find  the  value  of  J  •*•  (f  +f  )  -}.  Ans.  H- 
Parentheses  indicate  that  the  inclosed  operations  must  be  performed 

first.     For  example,  in  the  above,  -f  is  added  to  %  and  then  f  is  divided  by 
the  sum. 

68.  Find  the  value  of  2J-7S£+5j£-562-.  .4ns.  614£. 

69.  Simplify  («)*,  (b)  -?,   (c)  -    ,  (d)  -\l  Ans.   l{),  £,  H,  11*. 


30  PRACTICAL  MATHEMATICS 

70.  From    75  f,   take    12ft.  Ann.  62J?. 

71.  Multiply  21  -5-  i  by  }  of  (i+l).  Ant.  2^. 


72.  Find  the  value  of  •-•  4rw.  2|J. 


Suggestion.     First,  multiply  3?  by  8  J,  second,  4f  by  2^j,  then  divide  the 
first  product  by  the  second. 

73.  133-2ft-6A+3-l15a+8J-t?-10H»?  Ans.  4?*. 

74. 


(si-^-f)  X2 
75.  Evaluate  - 


_1.*.K 

4' 

Solution.  The  word  evaluate  means  that  the  indicated  operations 
should  be  performed  and  the  value  of  the  expression  found.  At  first 
decide  what  operations  must  be  performed  first,  what  second,  and  so 
on.  Then  do  these  operations  as  simply  as  possible. 


4 

76.  Evaluate  ^r^rSj^-  ^rw.  5»0. 


77.  Add  f  Xy  to  JX(4J-2J). 

D 

78.  Subtract  §  of  f  from  £  of  |.  vlrw.  /,. 

79.  f  of  20  is  ^  of  what  number?  Ans.  111. 

1      21     34      4 

80.  Find  the  simplest  expression  for  5T~q+2  ~4«"  '^nj*g  ^' 

Perform  the  operations  indicated  in  the  following  five  exercises: 
t    ,,  32    (41  +7j)  +81 


'       3I-2J 

83 
"• 


86. 


In  the  following  four  exercises,  the  letters  stand  for  values  as  follows: 
<i  =  14$,  6  =  16?,  c  =  33j,  and  d  =  27§;  find  the  values  of  the  fractions 
expressed  by  the  letters. 


86.         --.  Ans.   161.     88.  -  An.. 

c  —  o  — 


COMMON  FRACTIONS 


31 


90.  An  alloy  is  composed  of  92  Ib.  of  copper,  17  Ib.  of  zinc,  and  7  Ib. 
of  tin.     What  part  of  the  alloy  is  of  each  metal? 

Ans.  ||  copper;  ^^  zinc;  T|?  tin. 

91.  Three  men  did  a  piece  of  work  in  26 \  days  for  which  they  received 
$344£.     What  was  the  average  pay  per  day  for  each  man? 

Ans.  $4f 

29.  Pitch  and  lead  of  screw  threads. — The  pitch  of  a  screw 
thread  is  the  distance  from  the  center  of  the  top  of  one  thread 
to  the  center  of  the  top  of  the  next. 

The  lead  of  a  screw  thread  is  the  distance  the  screw  will 
move  forward  in  a  nut  for  each  complete  turn  of  the  screw. 


Single  Threaded 


H  --  L-  --H 


Double  Threaded 


Triple  Threaded 


L  =  Lead,     P  =  Pitch 


FIG.  2. 


For  a  single-threaded  screw  the  pitch  and  the  lead  arc 
equal;  but  for  a  double-threaded  screw,  the  lead  is  twice  the 
pitch;  and  for  a  triple-threaded  screw,  the  lead  is  three  times 
the  pitch. 

In  a  single-threaded  screw,  there  is  only  one  thread  run- 
ning around  the  screw;  in  a  double-threaded  screw  there  are 
two  threads  running  side  by  side  around  the  screw;  and  a 
triple-threaded  screw  has  three  threads  side  by  side  running 
around  the  screw. 

The  above  statements  are  made  clearer  by  reference  to 
the  figure. 


32 


PRACTICAL  MA  THEM  A  TICS 


92.  According  to  the  Franklin  Institute  standards  for  the  dimensions 
of  bolts  and  nuts,  a  |-in.  bolt  has  11  threads  per  inch.     What  is  the 
pitch?     The  lead  if  single-threaded?     How  many  full  turns  of  the  nut 
will  it  take  to  advance  the  bolt  2J  in.?  Ans.  jt  in.;  ft  in.;  24}. 

93.  A  4i-in.  bolt  has  2  j  threads  per  inch.     What  is  the  pitch?    What 
is  the  lead  if  triple-threaded?  Ans.  i*j  in.;  1ft  in. 

94.  In  a  special  threaded  screw  for  a  screw-power  stump  puller,  the 
screw  is  double-threaded  with  a  pitch  of  H  in-     How  many  turns  of  the 
nut  are  required  to  lift  the  stump  4J  ft.?  Ans.  37 ft. 

30.  The  micrometer. — The  screw  is  used  in  very  many 
mechanical  devices.  Many  of  these  will  be  used  in  illustrative 
problems  in  later  chapters.  The  use  of  the  screw  in  measuring 
small  distances  where  great  accuracy  is  required,  is  illustrated 
in  the  ordinary  micrometer  shown  in  Fig.  3. 


A  —  Frame 

B  —  Anvil 

C — Spindle  or  Screw 

D  —  Sleeve  or  Barrel 

E—  Thimble 


FIG.  3. — Micrometer. 

The  object  to  be  measured  is  placed  between  the  anvil,  B, 
and  the  spindle,  C.  The  spindle  has  a  thread  cut  40  to  the 
inch  on  the  part  inside  the  sleeve,  D.  The  thimble,  E,  is 
outside  the  sleeve  and  turns  the  spindle.  The  thimble  has  a 
beveled  end  that  is  divided  into  25  equal  divisions.  The  sleeve 
is  graduated  into  divisions  each  ^V  °f  an  inch,  every  fourth  of 
which  is  marked  1,  2,  3,  etc.  The  numbered  marks  then  rep- 
resent tenths  of  an  inch. 

If  the  thimble  is  turned  through  one  of  its  graduations,  the 
spindle  is  evidently  advanced  -5*5  of  ^V  in.  =  T^TV  in.  The 
vernier,  which  will  be  described  later  (Art.  167)  enables  one 
to  read  to  -jV  of  one  of  the  graduations  on  the  thimble  and 
hence  makes  it  possible  to  measure  -fa  of  TTfSv  in.  =TT^T)  in. 

The  readings  of  a  micrometer  are  usually  stated  in  decimals 
of  an  inch. 


COMMON  FRACTIONS 


33 


95.  Find  the  distance  the  spindle  advances  when  the  thimble  makes  7 
full  turns  and  17  divisions.  Ans.  t\fo8o  or  0.192  in. 

96.  What  is  the  measurement  when  the  reading  on  the  sleeve  is  5 
graduations  and  on  the  thimble  14?  Ans.  ^ftj-  or  0.139  in. 


31.  Screw  gearing.  —  Spiral  or  screw  gearing  is  often  used 
where  it  is  desired  to  reduce  the  speed  greatly.  The  teeth  are 
arranged  in  the  same  manner  as  the 
threads  of  a  screw.  A  screw  wheel  may 
have  one  tooth  or  any  number  of  t,eeth. 
A  one-toothed  wheel  corresponds  to  a 
one-threaded  screw,  a  many-toothed  wheel 
to  a  many-threaded  screw. 

In  Fig.  4,  the  upper  wheel  has  12  teeth, 
and  corresponds  to  a  12-threaded  screw; 
while  the  lower  wheel  has  36  teeth,  and 
corresponds  to  a  36-threaded  screw. 
Here  the  smaller  wheel  makes  three  revolu- 
tions while  the  larger  is  revolving  once. 
If  the  smaller  had  but  one  tooth  or  was 
single  threaded,  it  would  make  36  turns  to 
one  of  the  larger  wheel. 

When  the  number  of  threads  or  teeth  on  the  smaller  wheel 
is  few,  the  smaller  wheel  is  called  a  worm,  and  the  larger  the 
worm  wheel. 


FIG.    4.— Righthand 
spiral  gears. 


97.  If  a  worm  is  2-pitch  and  single-threaded,  how  many  inches  will 
it  cause  the  circumference  of  the  worm  wheel  to  advance  for  one  revolu- 
tion of  the  worm?     How  many  inches  if  double-threaded?     If  triple- 
threaded?     If  6-threaded?     (2-pitch  means  2  threads  to  the  inch.) 

98.  If  a  single-threaded  worm,  making  20  revolutions  per  second,  turns 
a  worm  wheel  having  54  teeth,  Fig.  5,  how  many  revolutions  per  minute 
will  the  worm  wheel  make?  Ans.  22 1. 

99.  In  turning  a  piece  in  a  turning  lathe  the  distance  the  cutting  tool 
advances  along  the  piece  for  each  revolution  is  called  the  feed.     The 
feed  is  usually  given  as  a  fraction  of  an  inch.     How  long  will  it  take  to 
turn  a  piece  2J  ft.  long,  if  it  makes  17  revolutions  per  minute,  and  the 
feed  is  ^  in.?  Ans.  28A  minutes. 

100.  How  many  turns  per  minute  is  a  piece  making  if  4  ft.  of  length 
is  turned  in  75  minutes  when  the  feed  is  -fa  in.?  Ans.  6||. 

101.  What  feed  is  necessary  to  run  a  cut  of  45  in.  in  10  minutes  at 
36  revolutions  per  minute?  Arts.  |  in. 

3 


34  PRACTICAL  MATHEMATICS 

102.  In  planing  an  aluminum  casting  of  width  10)  in.  and  length 
12}  in.,  find  the  time  required  if  cutting  speed  is  40  ft.  per  minute,  return 
speed  80  ft.  per  minute,  and  feed  /j  in.  Ans.  27  minutes  nearly. 

Siiggestion.  The  cutting  tool  cuts  a  strip  V*  in.  wide  each  time  across. 
Or  it  takes  64  times  across  to  plane  a  strip  one  inch  wide.  If  the  cut  is 


Lcfthand  single  thread.  Righthand  double  thread 

FIG.  5. — Worm  and  worm  wheel. 

made  the  long  way  of  the  casting  it  takes  64X10J  =672  cuts.  The  re- 
turn strokes  take  one-half  the  time  of  the  cutting  strokes.  No  allowance 
is  made  for  over  run  in  the  strokes. 

103.  A  A-in.  twist-drill  has  a  speed  of  130  revolutions  per  minute 
(R.  P.  M.)  when  cutting  steel.  How  long  will  it  take  to  drill  through  a 
f-in.  plate  if  120  revolutions  are  required  to  drill  1  in.? 

Ans.  \\  minutes. 


FIG.  6.— Twist  drill. 

104.  A  lA-in.  drill  makes  66  R.  P.  M.  in  iron  and  has  a  feed  of  sV  in. 
How  long  will  it  take  to  drill  20  holes  through  a  $-in.  plate  if  i  minute  is 
allowed  for  setting  for  each  hole?  Ans.  28i*r  minutes. 

105.  In  drilling  through  mild  steel  1  f  in.  thick,  a  hole  1  in.  in  diameter 
is  drilled  in  If  minutes;  find  the  distance  drilled  per  minute. 

Ans.  \\  in. 

106.  A  A-in.  drill  can  make  320  R.  P.  M.  in  brass.     If  120  turns  are 
made  to  drill  1  in.,  find  how  many  holes  can  be  drilled  in  a  i-in.  plate  in 
1  hour  if  J  the  time  is  used  in  setting  the  drill.  Ans.  192. 

107.  The  following  rule  is  often  used  to  find  the  weight  of  round  steel 
and  wrought  iron:  Square  the  diameter  in  inches  and  multiply  by  |, 


COMMON  FRACTIONS  35 

the  product  is  the  weight  in  pounds  of  1  ft.  of  the  bar.  (The  product  of  a 
number  multiplied  by  itself  is  the  square  of  the  number.)  Using  this 
rule  find  the  weights  of  round  bars  of  steel  of  the  following  dimensions: 

(1)  Diameter  f  in.,  length  12|  ft.  Ans.  25jf  Ib. 

(2)  Diameter  2f  in.,  length  2J  ft.  Ans.  41&  Ib. 

(3)  Diameter  8|  in.,  length  1|  ft.  Ans.  326 f  Ib. 

108.  Three  pipes  can  empty  a  reservoir  in  6,  5,  and  4  hours  respectively. 
How  long  will  it  take  them  to  empty  it  if  running  together? 

Ans.  If?  hours. 

Solution.  Since  the  first  pipe  can  empty  the  reservoir  in  6  hours,  it 
can  empty  f  of  the  reservoir  in  1  hour.  Reasoning  in  the  same  manner, 
the  second  pipe  can  empty  g  of  it  in  1  hour,  and  the  third  pipe  I  of  it  in 
1  hour.  Hence,  the  three  pipes  can  empty  £+i+l  =  i?  of  the  reservoir 
in  1  hour.  To  empty  the  whole  reservoir,  it  will  take  all  the  pipes  to- 
gether, l-r-|J=l|?  hours. 

109.  In  the  preceding  exercise,  how  long  will  it  take  to  empty  the 
reservoir  if  it  is  full  to  begin  with,  and  the  first  two  pipes  are  emptying 
out  of  and  the  third  emptying  into  the  reservoir?  Ans.  8|  hours. 

110.  Three  pipes  open  from  a  tank.     The  first  alone  can  empty  it 
in  6  hours,  the  second  in  4  hours,  and  the  third  in  3  hours.     How  many 
hours  will  it  take  to  empty  the  tank  if  the  pipes  are  all  running  together? 

Ans.  1|. 

111.  In  the  previous  exercise  the  tank  is  full  to  begin  with  and  the 
first  and  third  pipes  are  emptying  out  of,  and  the  second  emptying  into 
the  tank.     How  long  will  it  take  to  empty  the  tank?       Ans.  4  hours. 

112.  If  A  can  do  a  piece  of  work  in  9  hours  and  B  can  do  the  same  work 
in  6  hours,  how  long  will  it  take  them  if  working  together?     What  part 
can  each  do  in  1  hour?     What  part  of  the  work  can  both  together  do  in 
1  hour?  Ans.  3§  hours;  $  and  £;  -f$. 

113.  If  A  can  do  a  piece  of  work  in  10  J  days  and  B  can  do  the  same 
work  in  81  days,  what  part  of  the  work  can  they  both  do  in  one  day? 
In  how  many  days  can  they  do  the  work  if  working  together? 

Ans.  A;  4f. 

114.  If  one  gang  of  men  can  do  a  piece  of  work  in  20  days,  and  another 
gang  can  do  the  same  piece  of  work  in  25  days,  how  long  will  it  take  both 
gangs  working  together  to  do  the  work?  Ans.  11|  days. 

115.  A  contractor  is  to  grade  a  street  in  30  days.     Fifteen  men  work 
on  the  grading  for  20  days  and  do  one-half  of  the  work.     How  many 
men  must  work  for  the  next  10  days  to  finish  the  grading?      Ans.  30. 

116.  In  10  days,  15  men  do  -fa  of  a  piece  of  work.     How  many  men 
will  it  take  to  finish  the  work  in  15  days? 

117.  How  many  turns  must  be  made  with  a  triple-threaded  screw 
having  4j  threads  to  the  inch  to  have  it  advance  a  distance  of  3  in.? 

118.  The  lead  screw  on  the  table  of  a  milling  machine  has  a  double 
thread  with  a  pitch  of  J  in.     How  many  inches  per  minute  is  the  feed  if 
the  lead  screw  is  making  4  R.  P.  M.? 


36  PRACTICAL  MATHEMATICS 

119.  A  man  who  owns  $  of  a  city  block  valued  at  $140,000,  gold  11, 
of  his  sharp.     What  is  the  value  of  the  part  he  has  left? 

Ann.  $15,000. 

120.  A  merchant  bought  a  stock  of  goods  for  $7426.50  and  sold   \ 
of  it  at  an  advance  of  \  the  cost,  J  of  it  at  an  advance  of  i  the  cost,  and 
the  remainder  at  a  loss  of  1*5  the  cost.     Did  he  gain  or  lost*  and  how 
much?  Ana.  Gained  $1516.24. 

121.  A  sum  of  money  is  divided  among  four  persons.     The  first  ro- 
ceived  1*0  of  the  amount,  the  second  J,  the  third  J,  and  the  fourth  the 
remainder  which  is  $5000.     What  is  the  amount  each  received? 

Ans.  $6000;  $5000;  $4000;  $5000. 

122.  If  it  takes  4  tons  of  coal  to  heat  as  much  as  6  cords  of  wood, 
which  is  the  cheaper  if  coal  is  $7J  a  ton  and  wood  $5J  a  cord?     How 
much  more  will  one  cost  than  the  other  to  heat  a  house  that  requires  1 1 
tons  of  coal  a  year?  Ans.  Coal,  $lf. 

123.  Which  is  the  cheaper  and  how  much,  to  have  a  17J  cent  an  hour 
boy  take  13  J  hours  to  do  a  certain  piece  of  work,  or  have  a  42  §  cent  an 
hour  man  do  it  in  4$  hours?  Ans.  The  man  38}  cents  cheaper. 

124.  A  piece  of  work  when  forged  weighed  214$  Ib.     After  being 
turned  down  it  weighed  156f  Ib.     The  forging  cost  16i  cents  a  pound 
and  the  metal  turned  off  sold  at  3}  cents  a  pound.     Find  the  net  cost  of 
the  metal  in  the  finished  piece.  Ans.  $33.51^4. 

126.  A  machinist  drills  6  holes  through  a  piece  that  is  2J  in.  thick. 
The  drill  is  1J  in.  in  diameter  and  makes  154  R.  P.  M.  with  a  feed  of 
B'O  in.  How  many  minutes  does  it  take  if  3  minutes  are  used  in  setting 
for  each  hole?  Ans.  23fi. 

126.  A  hole  6f  in.  deep  is  drilled  with  a  1  J-m.  drill  making  126  R.  P.  M. 
What  feed  is  required  to  drill  the  hole  in  3}  minutes?         Ans.  fa  in. 

127.  To  change  from  Centigrade  thermometer  reading  to  Fahrenheit, 
the  following  formula  is  used:  F  =  gC+32°,  where  C  is  the  Centigrade 
reading  and  F  the  Fahrenheit.     (1)  Find  F  when  C=22i°.     (2)  Find 
F  when  C  is  720°.  Ana.  (1)  72i°5  (2)  1328°. 

128.  To  change  Fahrenheit  thermometer  reading  to  Centigrade,  the 
following  formula  is  used:  C  =  8(F  — 32°),  where  C  is  the  Centigrade 
reading  and  F  the  Fahrenheit.     (1)  Find  C  when  F=8H°.     (2)  Find 
C  when  F  =  1760°.  Ans.  (1)  27 J°;  (2)  960°. 

129.  In  inspecting  steamboilers,  the  following  formula  is  often  used : 

PRF 

t    —      — . 


TX% 

where  /  =  thickness  of  plate  in  inches, 

P=  steam  pressure  in  pounds  per  square  inch, 
R  =  radius  of  boiler  in  inches  (i  of  diameter), 
F  =  factor  of  safety, 

T  =  tensile  strength  of  boiler  plate  in  pounds, 
%  =  percentage  of  strength  in  joints. 


COMMON  FRACTIONS 


37 


Find  the  thickness  of  the  boiler  plate  which  should  be  used  for  a  boiler 
50  in.  in  diameter  to  carry  120  Ib.  of  pressure  if  the  tensile  strength  is 
(>0,000  Ib.  Use  50%  as  the  strength  of  the  joints  and  a  factor  of  safety 
of  6.  Ans.  |  in. 

PRF        120  X  25  X  6 


Solution,    t  = 


T  X  %       60000  X 


=  I 


130.  Find  the  thickness  of  the  boiler  plate  for  a  72-in.  boiler  to  carry  a 
pressure  of  90  Ib.  with  a  tensile  strength  of  60,000  Ib.     Use  50%  as  the 
joint  strength  and  a  factor  of  safety  of  6.  Ans.  f  in.  nearly. 

131.  The  following  formula  is  used  in  finding  the  diameter  of  a  steam- 
oiler 

2tT  X  % 


D   = 


PF 


where  D  =  diameter  of  boiler  in  inches, 
t  =  thickness  of  plate  in  inches, 
T=  tensile  strength  of  boiler  plate  in  pounds, 
P  =  steam  pressure  in  pounds  per  square  inch, 
F  —  factor  of  safety, 
%  =  percentage  of  strength  in  joints. 

Find  the  diameter  for  a  steamboiler  having  a  f-in.  plate,  allowing  50% 
for  strength  of  joints  and  a  factor  of  safety  of  6,  with  a  tensile  strength 
of  60,000  Ib.,  and  125  Ib.  pressure  per  square  inch.  Ans.  50  in. 


FIG.  7. 

132.  Find  the  number  of  lines  in  a  paper  of  38  pages,  two  columns 
to  the  page,  each  10  J  in.  long,  and  15  lines  in  2  in.     How  many  words 
if  they  average  11  words  to  the  line?     How  long  would  it  take  to  read 
such  a  paper  at  the  rate  of  170  words  a  minute? 

133.  The  distance  F  across  the  flats  in  a  bolt  head  or  nut,  either  a 
square  or  a  hexagon  (Fig.  7),  is  equal  to  1-|  times  the  diameter  of  the 
bolt  plus  |  in.     As  a  formula  this  is 


Test  the  widths  across  the  flats  in  the  following  table  taken  from  a 
manufacturer's  catalog: 


Diameter  of  bolt  D 

I 

5 
16 

f 

7 
16 

f 

3 

i 

ii 

1  * 
A8 

If 

2 

Width  across  flats  F 

i 

2 

3~2 

i¥ 

& 

1*' 

n 

i& 

1  i4 

1  16 

29 
T6 

2f 

Ql 
«*5 

5  0  2 


CHAPTER  III 
DECIMAL  FRACTIONS 

32.  Definition.— Fractions  that  have  10,  100,  1000,  etc.  for 
denominators  are  decimal  fractions. 

Thus,  -iVo,  iWo'o,  T^TJ  1888  are  decimal  fractions. 

In  writing  a  decimal  fraction  it  is  convenient  to  omit 
the  denominator,  and  indicate  what  it  is  by  placing  a  point 
(.),  called  a  decimal  point,  in  the  numerator  so  that  there 
shall  be  as  many  figures  to  the  right  of  this  point  as  there 
are  ciphers  in  the  denominator. 

Thus,  TVo  is  written  0.53;  fWA  =  0.3756;  TJJ0  =  0.076;  J888  =  4.326. 

In  such  numbers  as  0.53  and  0.3765,  the  cipher  is  printed 
at  the  left  of  the  decimal  point  for  clearness;  but  it  is  not 
necessary  and  is  often  omitted. 

It  is  to  be  noted  that  when  there  are  fewer  figures  in  the 
numerator  than  there  are  ciphers  in  the  denominator,  ciphers 
are  added  on  the  left  of  the  figures  to  make  the  required 
number. 

From  the  meaning  of  the  decimal  fraction,  it  is  seen  that 
the  misplacing  of  the  decimal  point  changes  the  meaning 
greatly.  For  each  place  it  is  moved  to  the  right,  the  value 
of  the  decimal  fraction  is  multiplied  by  10;  and  for  each  place 
it  is  moved  to  the  left,  the  value  is  divided  by  10. 

Thus,  2.75  becomes  27.5  when  the  point  is  moved  one  place  to  the 
right,  and  0.275  when  the  point  is  moved  one  place  to  the  left.  In  the 
first  case  2.75  is  multiplied  by  10,  and  in  the  second  case  it  is  divided  by 
10. 

It  is  well  to  recall  the  fact  that  when  we  have  a  number 
such  as  3333,  where  the  same  figure  is  used  throughout, 
the  values  expressed  by  the  threes  vary  greatly.  For  every 
place  a  three  is  moved  toward  the  left,  its  value  is  increased 
ten  times;  and  as  we  pass  from  left  toward  the  right,  each 
three  has  one-tenth  the  value  of  the  one  to  the  left  of  it. 
Since  the  above  relations  hold  when  we  pass  to  the  right  of 

38 


DECIMAL  FRACTIONS  39 

the  place  representing  units,  we  have  the  following  relative 
values  of  the  places: 


[  -^ 

02 

-C 

1- 

-c 
H 

X 

00 

^ 
-t— 

c 

c 

rr 

-a 

3 

A 

~ 

Thousands 

Hundreds 

Tens 
Units 
Decimal  poi: 

Tenths 

Hundredths 

Thousandth; 

Ten-thousan 

Hundred-thc 

Million  ths 

Ten-million  t 

Hundred-mil 

0000.00000000 

33.  Reading  numbers. — The  whole  number  23,676  is  read 
twenty-three  thousand  six  hundred  seventy-six.  It  should 
be  noticed  that  no  word  "and"  is  used  in  reading  a  whole 
number. 

A  decimal  is  read  like  a  whole  number  except  that  the  name 
of  the  right-hand  place  is  added. 

For  example,  the  number  0.7657  is  read,  seven  thousand  six  hundred 
fifty-seven  ten-thousandths. 

When  a  whole  number  and  a  decimal  fraction  are  written 
together  the  word  "and"  is  used  between  the  two  parts  in 
reading. 

Thus,  73.2658  is  read,  seventy-three  and  two  thousand  six  hundred 
fifty-eight  ten-thousandths. 

Where  one  person  is  reading  numbers  for  another  to  write, 
it  is  not  customary  to  proceed  in  the  above  manner. 

Thus,  the  number  23.6785  may  be  read  twenty-three,  point,  sixty- 
seven,  eighty-five.  Or  we  may  read  it,  two,  three,  point,  six,  seven, 
eight,  five. 

EXERCISES  10 

Write  the  following  in  figures : 

1.  Three  hundred  fifty-six  ten-thousandths. 

2.  Two  hundred  fifty-six  and  twenty-three  thousandths. 

3.  One  hundred  fifty-five  millionths. 

4.  Four  hundred  fifty-six  thousandths. 

5.  Four  hundred  and  fifty-six  thousandths. 

6.  Three  hundred  twenty-five  and  twenty-five  ten-thousandths. 


40  PRACTICAL  MATHEMATICS 

Read  the  following  : 

7.  23.402.  11.  1200.3604. 

8.  2003.203.  12.  10,101.2301. 

9.  0.4256.  13.  5867.0067. 
10.  4200  0056.  14.  10,0000001. 

34.  Reduction  of  a  common  fraction  to  a  decimal  fraction.- 
A  decimal  fraction  differs  from  a  common  fraction  only  in 
having  1  with  a  certain  number  of  ciphers  annexed  for  the 
denominator.  The  common  fraction  can  then  be  changed  to 
a  decimal  by  reducing  it  to  a  fraction  having  1  with  ciphers 
for  a  denominator. 

It  is  evident  from  the  method  of  reducing  a  common  fraction 
to  one  with  a  different  denominator,  that  a  common  fraction 
can  be  changed  to  a  decimal  only  when  its  denominator  is  con- 
tained an  exact  number  of  times  in  10,100,1000,  or  10000,  etc. 


Thus,  I  =iV  or  0.4,  and,  T98  =-f^o  or  0.5625,  but  ?  cannot  be  expressed 
exactly  as  a  decimal  because  7  is  not  exactly  contained  in  10,  100,  or 
JOOO,  etc. 

To  reduce  a  common  fraction  to  a  decimal  proceed  as 
follows: 

RULE.  Annex  ciphers  to  the  numerator  and  divide  by  the 
denominator.  Place  the  decimal  point  so  as  to  make  as  many 
decimal  places  in  the  result  as  there  were  ciphers  annexed. 

Thus,  |=0.875,  Process.     8)7.000 

0.875 
and  $=0.2857+  Process.     7)2.0000 

0.2857+ 

The  sign,  +,  placed  after  the  number  indicates  that  there 
are  still  other  figures  if  the  division  is  carried  further. 

A  common  fraction  in  its  lowest  terms  will  reduce  to  an 
exact  decimal  only  when  its  denominator  contains  no  other 
prime  factors  than  2  and  5. 

Thus,  88«  reduces  to  an  exact  decimal  for  64  is  made  up  of  2X2X2X 
2X2x2,  while  j7z  cannot  be  reduced  to  an  exact  decimal  for  its  de- 
nominator contains  the  factor  3. 

35.  Decimal  fraction  to  common  fraction.  —  To  change  a 
decimal  fraction  to  a  common  fraction  proceed  as  follows: 


COMMON  FRACTIONS  41 

RULE.  Replace  the  decimal  point  by  a  denominator  having 
1  and  as  many  ciphers  as  there  are  decimal  places  in  the  original 
fraction.  (See  Art.  32.) 

Thus,    2.  375  =f§?$,    which    may   be    written    as    a    mixed    number, 

9371     —  O  3 

''ItiTS'tf  —  •**• 

EXERCISES  11 

Reduce  the  following  to  decimals  : 

1.  ff-  2.  §.  3.  if. 

4.  l\\.  5.  21-6%.  6.  62|gfg. 

Reduce  the  following  to  common  fractions  or  mixed  numbers  in  their 
lowest  terms: 

7.  0.440.  8.  0.98.  9.  0.03125. 

10.  0.00096.  11.  14.06225.  12.  42.030125. 

13.  Reduce  the  following  decimals  of  an  inch  to  common  fractions 
in  their  lowest  terms:  0.375;  0.359375;  0.28125;  0.171875;  0.078125. 


14.  Express  the  following  in  their  simplest  common  fractional  form: 
3.04f;  0.00§;  0.28f;  0.714?;  0.484f;  0.87i     Ans. 

304  ?     A-7,?  a 
Suggestion.  3.04§  =          =  - 


15.  Change  the  following  per  cents  to  their  simplest  common  fractional 
forms:  87J%;  133*%;  f%;  185$%;  1.85f%;  2.21H%. 

Ans.  1;  |;  T^;  If  5  ?V<r;  dlft&for- 

16.  Tell  without  trial  which  of  the  following  common  fractions  will 
reduce  to  exact  decimals:  f;  ^;  T75;  J|;  /g;  ¥\;  ^;  ^  ;  f|;  f;  J||. 

17.  Change  the  following  decimals  of  an  inch  to  the  nearest  64ths  of 
an  inch:  0.394;  0.709;  1.416;  1.89. 


36.  Addition   of  decimals.  —  RULE.     Write  the  numbers  so 
that  the  decimal  points  are  under  each  other.     Add  as  in  whole 
numbers,  and  place  the  decimal  point  in  the  sum  under  the  other 
decimal  points. 

Example.     Add  36.036,  7.004,  0.00236,  427,  723.0026. 

36.036 
7.004 
0.00236 
427. 

723.0026 
1193.04496      Ans. 

37.  Subtraction   of   decimals.  —  RULE.     Write  the  numbers 
so  that  the  decimal  points  are  under  each  other;  subtract  as  in 


42  PRACTICAL  MATHEMATICS 

whole  numbers,  and  place  the  decimal  point  of  the  remainder 
under  the  other  decimal  points. 

Example.     Subtract  46.8324  from  437.421. 

437.421 
46.8324 

390.5886     Ans. 

38.  Multiplication  of  decimals. — RULE.  Multiply  as  in 
whole  numbers,  and  point  off  as  many  decimal  places  in  the 
product  as  the  sum  of  the  numbers  of  the  places  in  the  factors. 

Example  1.  Example  2. 

Multiply  7.32  by  0.032.     Multiply  0.00264  by  0.000314. 
7.32  0.00264 

0.032  0.000314 

1464  1056 

2196  264 

0.23424      Ans.  792 

0.00000082896     Ans. 

Multiplying  a  whole  number  or  a  decimal  by  0.1  moves  the 
decimal  point  one  place  to  the  left;  by  0.01,  two  places;  by 
0.001,  three  places;  etc.  If  it  is  necessary,  zeros  are  prefixed 
to  the  multiplicand. 

Thus,  32.4  X  0.0001  =  0.00324. 

Multiplying  by  10,  100,  1000,  etc.,  moves  the  decimal  point 
1,  2,  3,  etc.,  places  to  the  right. 

39.  Division  of  decimals. — RULE.  //  the  number  of  decimal 
places  in  the  dividend  is  kss  than  the  number  in  the  divisor,  annex 
ciphers  to  the  dividend  till  there  are  as  many  or  more  decimal 
places  as  in  the  divisor.  Divide  as  in  whole  numbers,  and  point 
off  as  many  decimal  places  in  the  quotient  as  there  are  more 
decimal  places  in  the  dividend  than  in  the  divisor. 

Example  1.  Example  2. 

Divide  0.4375  by  0.125.  Divide  4365  by  0.005. 

0.43750.125  0.005)4365.000 
375 


3.5      Ans.  873.000       Ans. 


625 
625 


Dividing  by  0.1,  0.01,  0.001,  etc.,  moves  the  decimal  point 
1,  2,  3,  etc.,  places  to  the  right.  Dividing  by  10,  100,  1000, 
etc.,  moves  the  decimal  point  1,  2,  3,  etc.,  places  to  the  left. 


DECIMAL  FRACTIONS  43 

40.  Accuracy  of  results.  —  Often  we  are  asked  to  give  a  result 
correct  to  a  certain  number  of  decimal  places.  Thus,  if  in 
working  a  problem  we  have  a  result  as  47.264735,  and  wish  to 
write  it  correct  to  three  places,  it  is  47.265  —  .  Correct  to  two 
places,  it  is  47.26+,  correct  to  one  place,  47.3  —  ,  correct  to 
five  places  47.26474-. 

The  last  place  taken  is  written  one  larger  when  the  next 
figure  to  the  right  is  5  or  more. 

The  part  to  the  right  of  the  last  place  taken  is  thrown  away 
when  the  first  figure  of  it  is  less  than  5. 

In  this  way  we  call  a  half  or  more  of  the  last  unit  taken,  a 
whole  one  of  those  units,  and  throw  away  anything  less  than 
a  half. 

The  sign,  +,  is  used  to  show  that  the  accurate  result  is 
larger  than  the  one  given,  that  is,  that  something  has  been 
thrown  away;  and  the  sign,  —  ,  is  used  to  show  that  the  accu- 
rate result  is  smaller  than  the  one  given,  that  is,  that  something 
has  been  added. 

EXERCISES  12 

Add  up  and  test  by  adding  down  : 

1.  864.2                          2.       5.82  3.  49.235 

43.276                                   .486  86.426 

21.004                               41.987  92.784 

9824.246                             987.  46.324 

47.02                               201.478  33.867 

39.09                               804.008  99.847 

34.396 


4.  18i  +  17f  +29^+14.672+34?  =  ?  Ans.   114.791+. 

5.  87.46  \  +93.27^+43.2906  +0.0047  +  17£  =  ?        Ans.  241.2027-. 

6.  8.706  +  7.898+43|+89A+14§  =?  Ans.   163.833  +  . 
7.1-0.640726  =  ?    '  Arts.  0.359274. 

8.  2-1.798642  =  ?  Ans.  0.201358. 

9.  4.8728-0.987  =  ?  Ans.  3.8858. 

10.  75.0075-2.75903  =  ?  Ans.  72.24847. 

11.  470.84-86.4396  =  ?  Ans.  390.4004. 

12.  From  one  thousand  take  five-thousandths.  Ans.  999.995. 

13.  From  three  million  and  one-millionth  take  one-tenth. 

Ans.  2999999.900001. 

14.  78.896-53.5987  =  ?  Ans.  25.2973. 

15.  81.35-11.678956  =  ?  Aws.  69.671044. 

16.  From  nine  hundred  nine  take  nine  hundred  and  nine  thousandths. 

Ans.  8.991. 


44 


PR  A  CTICAL  MA  THEM  A  TICS 


17.  From  37.75J  take  4.43f.  Ana.  33.32126. 

18.  One  quart  liquid  measure  has  57.75  cu.  in.,  and  1  quart  dry  mca*- 
ure  has  67.200025  cu.  in.     How  many  cubic  inches  larger  is  the  dry 


quart  than  the  liquid  quart? 

19.  3.62X0.0037  =  ? 

20.  2.53X0.00635  =  ? 

21.  0.00076X0.0015  =  ? 

22.  7.789X4.924  =  ? 

23.  2.236X799  =  ? 

24.  2.967X2.967  =  ? 

25.  8.943X1§  =  ? 

Why  would  it  be  best  not  to  reduce 
1J  to  a  decimal  before  multiplying? 

The  multiplication  can  be  carried  out 
readily  as  shown  here,  the  process  is  as  if 
the  multiplicand  were  a  whole  number. 


Ans.  9.460626. 

Ans.  0.013394. 

Ant.  0.0160655. 

Ans.  0.00000114. 

Ans.  38.353+. 

Ans.  1786.564. 

Ans.  8.803  +. 

Ans.  14.905. 

Process. 


8.943 


2981 
5962 
8  943 


14.905 


=  J  of  8943 
=  §  of  8943 
= 1  X  8943 
=  1§X8.943 


Ans.  23.218-. 

Ans.  2.1657  +  . 

Ans.  1499.2176 -. 

Ans.  0.000105+. 

Ans.  0.9745  +  . 

Ans.  2.7781 +. 


26.  2.43|X9.5?  =  ? 

27.  0.0439X491  =  ? 

28.  1X3.1416X7.1X7.1X7.1  =  ? 

29.  9.3X0.0042X0.0027  =  ? 

30.  49367  X  0. 002 1  X  0. 0094  =  ? 

31.  0.81X91X0.3375  =  ? 

32.  Multiply  3f  thousandths  by  3f  hundredths. 

Ans.  0. 0001406  +. 

33.  1  kilogram  equals  2.2046  pounds;  how  many  pounds  in  275.3 
kilograms?  .4ns.  606. 926+. 

34.  Multiply  each  of  the  following  numbers  by  0.1,  0.01,  0.001,  0.0001, 
10,  100,  1000,  10,000:  94,  47.368,  0.023,  3.42. 

36.  67.56785 -=-0.035  =  ?  Ans.  1930.51. 

36.  0.567891-i-8.2  =  ?  Ans.  0.069255. 

37.  Divide  43.769  by  4.76  correct  to  four  places. 

Process. 
43.7690004.70 
4284 


9.1951 


Explanation.  Since  the  quotient  is  to 
be  correct  to  four  places,  the  dividend 
must  contain  four  more  decimal  places 
than  the  divisor.  Three  zeros  are  added 
to  make  this  number.  Since  the  fifth 
decimal  figure  in  the  quotient  is  not  less 
than  5.  the  answer  is  9.1952  —  . 


929 

476  An*. 
4530 

1JM 

2460 


is  9.1 952-. 


800 
476 
324 


DECIMAL  FRACTIONS 


45 


In  the  next  5  exercises,  find  the  result  correct  to  four  decimal  places. 

38.  9.375+4.76  =  ?  Ans.  1.9695  +  . 

39.  89.7201  +  3.276  =  ?  Ans.  27.3871-. 

40.  34.675+4.375  =  ?  Ans.  7.9257  +  . 

41.  43.45+3.1416  =  ?  Ans.  13.8305+. 

42.  3.1416  +  6.67  =  ?  .4ns.  0.4710  +  . 

43.  Divide  324.8  by  4000. 

Explanation.  Cancel  the  zeros  in  the  divisor. 
Since  this  divides  the  divisor  by  1000,  the  dividend 
must  be  divided  by  1000,  this  is  done  by  moving  the 
decimal  point  three  places  to  the  left. 

44.  (a)  25  +  500  (b)  2+2000 

(d)  1.44  +  12,000       (e)  3.075  +  5000 

45.  (a)   1+0.0001  (b)  0.66+0.011 
(d)  0.00072+8           (e)   6600  +  0.0022 

46.  Divide  3.1416X1.25X50  by  0.8X2.75X3. 
Explanation.     The  cancelling  may 

be  done  as  in  whole  numbers,  paying 
no  attention  to  the  decimal  point. 
When  through,  point  off  as  many 
places  in  the  result  as  the  difference 

between  the  sum  of  those  above  and  0.$X2./75X    3 

the    sum    of    those    below    the  line.  tf 

Thus,    in   the  example  there  are  six 

places  above  and  three  below  the  line;  hence,  the  result  has  three  decimal 
places.     If  there  are  more  places  below  than  above,  add  ciphers  until 
there  are  as  many  above  as  below. 
Find  the  value  of  the  following: 
2.75X46.2X100 


Process. 
4p0p)0.3248 
0.0812 

(c)  9009  +  11,000 
Cf)   5684  +  14,000 
(c)  525+0.025 
(/)  3.03+0.03 

Process. 


=  29.750. 


47. 


48. 


49. 


60. 


61. 


to  units. 


2.5X2.8 
0.7854X5X5X8 

1.25 
6.4X0.84X9.6X1.44 

8X9.2X1.28 
0.7854X6X6X12.5X1728 

231X31.5 
8.5X9.25X3.66X1728 


to  three  places. 

to  three  places. 

to  two  places. 


Ans.  1815. 


Ans.  125.664. 


Ans.  0.789-. 


Ans.  83.93  +  . 


to  two  places. 


2150.42 

52.  What  is  the  inside  diameter  of  a  pipe 
which  is  7.34  in.  outside  diameter  and  is  made 
of  iron  0.743  in.  thick?  Ans.  5.854  in. 

63.  A  hundred  pounds  of  coke  were  found 
to  contain  5.79  Ib.  of  ash  and  0.597  Ib.  of 
sulphur,  the  rest  was  carbon.  How  much 
carbon  was  there?  Ans.  93.613  Ib. 

54.  A  steam  pump  delivers  26.44  gallons 
per  stroke.  A  gallon  of  water  weighs  8.355 
Ib.  What  weight  of  water  will  it  deliver  in 
117  strokes?  Ans.  25,846  Ib. 


Ans.  231.24-. 


0.743 


FIG. 


8.— Cross 
of  pipe. 


46  PRACTICAL  MATHEMATICS 

65.  In  1  Ib.  of  phosphor  bronze  0.925  is  copper,  0.07  w  tin,  and  0.00.r> 
is  phosphorus.     How  much  of  each  is  there  in  369.523  pounds  of  phosphor 
bronze?  Ans.  34 1. 8088 -;  25.8666  +  ;  1.8476+  Ib. 

66.  A  certain  paper  weighs  68  Ib.  per  ream  of  500  sheets.     If  tin- 
paper  costs  5J  cents  per  pound,  how  much  will  2fg  reams  cost? 

Ans.  $8.41. 

67.  Manganese  bronze  contains  the  following:  copper  0.89,  tin  0.1, 
manganese  0.01.     How  much  of  each  metal  is  there  in  a  propeller  weigh- 
ing 2378J  Ib.?  Ans.  2117.16  +  ;  237.88  +  ;  23.79-  Ib. 

58.  Add  •—  and  _*>  divide  the  result  by  7Jf,  and  change  the  quotient 

OITT          «t 
to  a  decimal.  An*.  0.125. 

69.  From  £i  X2J  subtract  the  product  of  0.075  and  Ij,  divide  tho  re- 
mainder by  12,  and  change  the  result  to  a  decimal.  Ans.  0.0375. 

cn    Q.      ...     (3.2+0. 004  -1.11DX0.25 

*'  Simpllfy  (4  +0.2)  -17.907 

61.  Simplify  (1§+H  -0.024)  -=-(15* -1.209).  Ans.  0.214 +. 

62.  Simplify  (i±M>  XOOOO^  An*.  0.00051 28+. 

U.U7O 

...     (3.71-1. 908)  X7.03 

63.  Simplify — ~ —    —  •  Ans.  6.405+. 

—  j'ss 

...     (201  +2.25 X0.004)  -=-  (1.0337 -31.09 X0.03) 

64.  Simplify  -  _____ 

Ans.  424,573.5-. 

66.  How  many  lengths  each  0.0275  of  a  foot  are  contained  in  27.2375 
ft.?  Ans.  990.45+. 

66.  The  expenditures  of  the  British  Naval  Service  were  as  follows: 
for  the  year  1906-7,   £31,472,087;  for  the  year  1907-8,   £31,251,156; 
express  these  sums  to  the  nearest  dollar  if  £1  =$4.8665. 

Ans.  $153,158,911;  $152,083,751. 

67.  A  wood  dealer  charged  $33.62  for  a  pile  of  wood  containing  7J 
cords.     What  error  did  the  dealer  make  if  wood  was  worth  $4.25  a  cord? 

Ans.  $0.68. 

68.  If  a  circle  is  3.1416  times  as  far  around  as  through  it,  find  the 
number  of  feet  around  a  cart  wheel  3.75  ft.  through.     Find  the  distance 
around  the  earth  if  the  diameter  is  7918  miles. 

Ans.  11.781;  24,875+  miles. 

69.  Find  the  value  of: 

27,750  shingles  at  $4.25  per  thousand.  Ans.  $117.94. 

47,256  ft.  of  lumber  at  $45  per  thousand.  Ans.  $2126.52. 

126,450  bricks  at  $7.75  per  thousand.  Ans.  $979.99. 

45,350  ft.  of  gas  at  $0.85  per  thousand.  Ans.  $38.55. 

70.  What  is  the  cost  of  carbon-steel  rails  to  lay  6  miles  of  street  car 
track,  if  the  rails  weigh  129  Ib.  per  yard,  and  cost  $28  per  ton.     (1  mile  = 
1760yd.)  Ans.  $38,142.72. 

71.  It  has  been  determined  by  experiment  that  each  square  foot  of 
steam  radiation  will  give  off  to  the  surrounding  air  about  3  heat  units 


DECIMAL  FRACTIONS  47 

per  hour  per  degree  difference  between  the  air  in  the  room  and  the  steam 
radiator.  If  the  temperature  of  the  radiator  is  212°  and  that  of  the  room 
70°,  how  many  heat  units  will  be  given  off  per  hour  on  24,000  sq.  ft.  of 
radiating  surface?  How  many  pounds  of  coal  will  it  take  to  make  this 
steam  if  1  Ib.  of  coal  contains  10,000  heat  units? 

Ans.  10,224,000;  1022.4. 

72.  Nickel  steel  will  stand  a  pull  of  about  90,000  Ib.  per  square  inch 
in  cross  section.     What  pull  will  a  bar  0.786  in.  wide  and  0.237  in.  thick 
stand?  Ans.   16,765+  Ib. 

Suggestion.  The  area  of  the  cross  section  is  found  by  multiplying 
0.786  by  0.237. 

73.  The  composition  of  white  metal  as  used  in  the  Navy  Department 
is  as  follows:  tin  7.6  parts,  copper  2.3  parts,  zinc  83.3  parts,  antimony 
3.8  parts,  and  lead  30  parts.     Find  the  number  of  pounds  of  each  in 
635  Ib.  of  the  white  metal. 

Ans.  tin  38;  copper  11.5;  zinc  416.5;  antimony  19;  lead  150. 
Suggestion.     Adding  7.6+2.3  +83.3  +3.8+30  =  127  =  whole  number  of 
parts. 

635 -j-127  =5  =  number  of  pounds  for  each  part. 
Multiply  5  Ib.  by  number  of  parts  of  each. 

74.  In  1909  the  number  of  horses  in  the  United  States  was  20,640,000. 
They  were  valued  at  $1,974,052,000.     Find  the  average  price  per  head. 

Ans.  $95.64+. 

76.  A  reamer  that  is  6  in.  long  is  1.2755  in.  in  diameter  at  the  small 
end,  and  1.4375  in.  at  the  larger  end.  Find  the  taper  per  foot.  (The 
taper  per  foot  means  the  decrease  in  diameter  per  foot  of  length.) 

Ans.  0.324  in. 

76.  A  creditor  receives  $0.76  on  each  dollar  due  him.     If  he  loses 
$326.40,  how  much  was  due  him?     What  would  he  have  received  if 
$7642  had  been  due  him?  Ans.  $1360;  $5807.92. 

77.  It  cost,  for  labor  and  materials,  $38,692.38  to  construct  7500  ft. 
of  car  track.     What  was  the  average  cost  per  foot?     What  would  be  the 
cost  for  100  miles  at  the  same  rate?     (1  mile  =  5280  ft.) 

Ans.  $5.159-;  $2,723,943.55. 

78.  During  1908  the  Daly- West  mine  marketed  12,760  tons  of  crude 
ore;  containing  2,683,830  Ib.  of  lead;  343,376  Ib.  of  copper;  454,149  oz. 
of  silver;  and  441.86  oz.  of  gold.     Find  its  value  if  lead  is  worth  3.925 
cents  per  pound;  copper  13.208  cents  per  pound;  silver  52.864  cents  per 
ounce;  and  gold  $18.842  per  ounce.  Ans.  $399,100.28. 

79.  One  cubic  foot  of  water  weighs  62.5  Ib. ;  find  the  volume  of  1  Ib.  of 
water.     Of  23  Ib.  Ans.  0.016  cu.  ft.;  0.368  cu.  ft. 

80.  One  cubic  foot  of  ice  weighs  57.5  Ib. ;  find  the  volume  of  1  Ib.  of  ice. 
Of  49.3  Ib.  Ans.  0.0174-  cu.  ft.;  0.857+  cu.  ft. 

81.  How  many  times  as  heavy  as  ice  is  water?     How  many  times  as 
heavy  as  water  is  ice?  Ans.   1.087  —  ;  0.92. 

82.  If  the  fire  under  a  steam  boiler  requires  3  Ib.  of  coal  per  horse- 
power per  hour,  find  the  cost  of  coal  at  $3.75  a  ton  to  run  a  160  horse- 
power boiler  for  30  days  of  10  hours  each.  Ans.  $270. 


48  PRACTICAL  MATHEMATICS 

83.  A  column  of  water  2.302  ft.  high  gives  a  pressure  at  the  base  of  1 
Ib.  per  square  inch.     Find  the  height  of  a  column  of  water  to  give  a 
pressure  of  256.3  Ib.  per  square  inch.     Find  the  pressure  per  square  inch 
of  a  column  of  water  1  ft.  high.     A  column  237.4  ft.  high. 

Ans.  590.0+  ft.;  0.4344+  Ib.;  103.1+  Ib. 

84.  The   Auditorium    Building   in    Chicago   has   a   cubic  content  of 
9,128,744  cu.  ft.,  and  cost  36  cents  per  cubic  foot.     Find  the  total  cost. 

Ans.  $3,286,347.84. 

86.  Using  U.  S.  standard,  the  gage  and  thickness  for  sheet  steel  is  as 
follows:  No.  00,  0.34375  in.;  No.  2,  0.265625  in.;  No.  4,  0.234375  in.; 
No.  7,  0.1875  in.;  No.  13,  0.09375  in.;  No.  28,  0.015625  in.  Find  the 
approximate  thickness  of  each  in  a  common  fraction  of  an  inch  having 
8,  16,  32,  or  64  for  a  denominator.  Ans.  J$,  JJ,  f},  -fa,  -fj,  ^j. 

86.  If  it  costs  $106.50  per  day  to  run  a  gang  of  men  and  a  rock  crusher 
giving  a  daily  output  of  200  cu.  yd.  of  crushed  rock,  find  the  cost  per 
cubic  yard.  Ans.  $0.5325. 

87.  In  building  a  certain  canal  lock  2HO  cu.  yd.  of  concrete  were  used. 
It  cost  $1.77  a  cubic  yard  for  mixing  and  placing  the  concrete.     The 
material  for  the  concrete  was  as  follows: 

3010  barrels  of  cement,  at  $3.02, 
1377  cu.  yd.  of  broken  stone  at  $1.37, 
393  cu.  yd.  of  screened  pebbles  at  $0.90, 
459  cu.  yd.  of  gravel  at  $0.67, 
500  cu.  yd.  of  sand  at  $1.78. 
Find  the  cost  of  the  concrete  work.  Ans.  $16,315.72. 

88.  In  building  a  concrete  viaduct  containing  2111  cu.  yd.,  the  total 
cost  was  1908  barrels  of  cement  at  $1.60;  1105  cu.  yd.  of  sand  at  $1.95; 
1468  cu.  yd.  of  stone  at  $1.48;  lumber  for  forms  $1140;  tools,  hardware, 
etc.,   $527.75;  water   $63.00;  labor  $7262.     Find  the  average  cost  per 
cubic  yard  of  concrete.  Ans.  $7.756  + 

89.  Number  8  (B.  &  S.)  gage  sheet  steel  is  0.1285  in.  thick  and  weighs 
5.22  Ib.  per  square  foot.     (1)  Find  the  thickness  of  a  pile  of  56  such 
sheets.     (2)  Find  the  nearest  whole  number  of  sheets  to  make  a  pile  1 
ft.  thick.     (3)  Find  the  weight  of  this  number  of  sheets  if  each  sheet  has 
4  sq.  ft.  Ans.  7.196  in.;  93;  1941.8+  Ib. 

90.  Number  25  (B.  &  S.)  gage  sheet  copper  is  0.0179  in.  thick  and 
weighs  0.811  Ib.  per  square  foot.     Answer  the  same  questions  as  in 
exercise  91.  Ans.  1.0024  in.;  670;  2173.5-  Ib. 

91.  An  iron  chain  made  of  l|-in.  round  iron  has  a  breaking  strain  of 
88,301  Ib.     If  the  chain  weighs  17.5  Ib.  per  foot,  how  long  would  the 
chain  have  to  be  to  break  of  its  own  weight  if  suspended  from  one  end? 

Ans.  5046-  ft. 

92.  Answer  the  same  question  as  in  exercise  93  for  a. chain  made  of 
-fi-in.  round  iron,  the  chain  weighing  0.904  Ib.  per  foot  and  breaking 
under  a  strain  of  4794  Ib.  Ans.  5303+  ft. 

93.  How  much  must  be  paid  for  1600  ft.  of  9teel  bar  weighing  1.87  11>. 
I>er  foot  and  costing  $4.65  per  hundred  pounds?  Am.  $139.13. 


DECIMAL  FRACTIONS 


49 


94.  If  a  steel  tape  expands  0.00016  in.  for  every  inch  when  heated, 
how  much  will  a  tape  100  ft.  long  expand? 

95.  A  round  piece  of  work  being  turned  in  a  lathe  is  1.4275  in.  in 
diameter.     What  is  the  diameter  after  a  cut  -fa  in.  deep  is  taken  in  the 
work?  Ans.  1.39625  in. 

96.  The  inside  diameter  of  a  steam  cylinder  before  boring  was  26  in. 
The  diameter  after  boring  was  26.3125  in.     How  deep  a  cut  was  taken  in 
boring?     In  boring,  20  turns  were  made  in  a  minute.     How  long  would 
it  take  to  bore  a  distance  of  28  in.  if  the  feed  was  0.0625  in.? 

Ans.  -£2  in.;  22.4  minutes. 

97.  Under  a  load  of  325  Ib.  a  wire  112  in.  long  and  0.09074  in.  in 
diameter  lengthened  0.265  in.     What  was  the  stretch  per  foot  to  four 
decimal  places?  Ans.  0.0284  in. 

98.  In  mixing  a  quantity  of  concrete  using  1  part  Portland  cement, 
2£  parts  of  sand,  3  parts  of  gravel,  and  5  parts  of  broken  stone,  it  was 
found  that  1  barrel  of  cement  averaged  1.18  cu.  yd.  of  concrete.     If  a 
barrel  of  cement  contains  3f  cu.  ft.,  how  many  cubic  feet  of  material  were 
put  into  one  cubic  yard  of  concrete?     Explain  how  this  could  be. 

Ans.  36.55  cu.  ft. 


FIG.  9. 

41.  Proportions  of  machine  screw  heads.  A.  S.  M.  E. 
standard. — In  the  following  four  problems  are  given  the 
four  standard  heads.  The  proportions  are  based  on  and 
include  the  diameter  of  the  screw,  diameter  and  thickness 
of  the  head,  width  and  depth  of  the  slot,  radius  for  round 
and  fillister  heads,  and  included  angle  of  flat-headed  screw. 

( 1 )  Oval  fillister  head  machine  screws. 
A  =  diameter  of  body. 

jB  =  1.64A  —  0.009=  diameter  of  head  and  radius  of  oval. 
C  =0.66A  -  0.002  =height  of  side. 
D=0.173A +0.015=  width  of  slot. 
E  =  %F  =  depth  of  slot. 
F  =0.1348+ C  =  height  of  head. 


f>o 


I'RA  CTICAL  MA  Til  EM  A  TICS 


99.  Given  the  values  of  A  find  those  of  B,  C,  D,  E,  and  F. 

A  B  C  D  E  F 

0.216   0.3452  0.1406  0.052   0.093   0.1868 

0.398   0.6437  0.2607  0.084   0  173   0.3469 

0.450   0.729   0.295   0.093   0.196   0.3927 


Fie;.   10. 


FIG.  11. 


Suggestion.     The  values  of  B,  C,  D,  E,  and  F  are  to  be  found  from  the 
given  value  of  A. 

5  =  1.64X0.216-0.009  =  0.3452. 
C  =  0.66  X0.216  -0.002  =0.1406. 

(2)  Flat  fillister  head  machine  screws. 
A  —  diameter  of  body. 

B  =  1.64A   -0.009  =diameter  of  head. 
C  =0.66A   -0.002  =height  of  head. 
£>=0.173A  +0.015  =width  of  slot. 
E  =  \C=  depth  of  slot. 

100.  Given  the  values  of  A  find  those  of  B,  C,  D,  and  E. 

A  B                 C                 D                 E 

0.112  0.1747  0.0719  0.034  0.036 

0.177  0.2813  0.1148  0.046  0.057 

0.320  0.5158  0.2092  0.070  0.105 

(3)  Flat  head  machine  screws. 
A  =  diameter  of  body. 

B  =2 A  -0.008  =  diameter  of  head. 

A  -0.008 
C= — i  739~  =depth  of  head. 

D  =  0.173A +0.015  =width  of  slot. 
#  =  jC  =  depth  of  slot. 

101.  Given  the  values  of  A  find  those  of  B,  C,  D,  and  E. 

A                 B                 C                 D  E 

0.086  0.164  0.045  0.030  0.015 

0.242  0.476  0.135  0.057  0.045 

0.372  0.736  0.209  0.079  0.070 


DECIMAL  FRACTIONS 


51 


(4)  Round  head  machine  screws. 
A  =  diameter  of  body. 
B  =  1.85 A  -0.005  =  diameter  of  head. 
C  =  0.7A  =  height  of  head. 
D=0.173A +0.015=  width  of  slot. 
=  depth  of  slot. 


102.  Given  the  values  of  A  find  those  of  B,  C,  D,  and  E. 

A                  B                  C  D                  E 

0.073           0.130           0.051  0.028           0.035 

0.164           0.298           0.115  0.043           0.067 

0  398           0.731           0.279  0.084           0.149 

103.  The  formula  for  determining  the  number  of  threads  per  inch  on 
machine  screws  is 

v  =       6-5 
A  +0.02 

where  N  is  the  number  of  threads  per  inch,  and  A  the  diameter  of  the. 
screw. 

Compute  the  number  of  threads  per  inch  for  machine  screws  of  the 
following  diameters:  0.242,  0.398,  0.450,  0.563,  0.750.  In  each  casR 
give  the  answer  to  the  nearest  whole  number. 

Ans.  25;  16;  14;  11;  8. 


CHAPTER  IV 
SHORT  METHODS  AND  CHECKS 

42.  Contracted  methods  and  approximate  results. — As 
a  rule  the  practical  man  does  not  need  a  large  number  of 
decimal  places.  The  results  of  all  measurements  are  at  best 
only  an  approximation  of  the  truth.  The  accuracy  depends 
upon  the  instruments,  the  method  used,  and  upon  the  thing 
measured.  All  that  is  necessary  is  to  be  sure  that  the  magni- 
tude of  the  error  is  small  compared  with  the  quantity  meas- 
ured. It  is  clear  that  in  a  dimension  of  several  feet,  a 
fraction  of  an  inch  would  probably  not  make  much  difference ; 
but  if  the  dimension  was  small,  such  an  error  could  not  be 
allowed. 

The  man  in  practical  work  uses  instruments  which  are  of 
such  accuracy  as  to  secure  results  suitable  to  his  purpose.  If 
he  requires  measurements  accurate  to  0.001  in.,  it  is  not  neces- 
sary for  him  in  a  computation  to  carry  his  work  to  0.00001 
in.  A  good  rule  to  go  by  is  not  to  calculate  to  more  than 
one  more  decimal  place  than  measurements  are  made. 

Thus,  if  a  measurement  of  3.265  in.  is  made,  and  it  is  to 
be  multiplied  by  3.1416,  it  is  not  necessary  to  multiply  in 
the  usual  way,  as  then  there  would  be  seven  decimal  places, 
while  the  measurement  was  accurate  to  only  three  places. 
If  the  multiplication  is  performed  by  multiplying  first  by 
the  left-hand  figure  of  the  multiplier,  and  then  passing  toward 
the  right,  we  have  the  following  forms: 

Form  in  full.  Contracted  form. 

3.265  3.265 

3.1416  3.1416 

9795  9795 

3265  3265 

13060  1306 

3265  32 

19590  19 


10.2573240  10.2572 

52 


SHORT  METHODS  AND  CHECKS  53 

It  will  be  noticed  that  one  more  decimal  place  is  retained 
than  the  desired  number. 

In  a  similar  manner,  division  can  be  contracted.  Suppose 
it  is  required  to  divide  0.04267  by  3.278,  and  secure  an  answer 
correct  to  four  significant  figures. 

The  division  in  the  full  and  contracted  forms  is  as  follows: 

0.042670000,  3.278  0.042670    3.278 

3278          0.013017  3278 


0.013017 


9890  9890 

9834_  _9834 

~5600  56 

3278  32 

23220  24 

22946  22 

274  2 

Hence  the  result  correct  to  four  significant  figures  is 
0.01302- . 

It  is  easy  to  follow  the  method  in  obtaining  the  above,  but 
it  is  hardly  worth  spending  time  upon  unless  one  is  to  do  much 
computing  of  this  kind. 

EXERCISES  13 

Solve  the  following  by  contracted  forms: 

1.  3.14159X3.14159  correct  to  four  decimal  places. 

Ans.  9.8696. 

2.  9,376,245-^3724  correct  to  the  unit's  place.  Ans.  2518. 

3.  1 00  H-3. 14159  correct  to  0.01.  Ans.  31.83. 

4.  87,659,734-^5467  correct  to  five  significant  figures. 

Ans.   16,034. 

5.  45.8636X26.4356  correct  to  five  significant  figures. 

Ans.   1212.4. 

6.  6.234X0.05473  correct  to  four  significant  figures. 

Ans.  0.3412. 

7.  4.326X0.003457  correct  to  five  significant  figures. 

Ans.  0.014955. 

43.  Other  methods. — -Numerous  short  methods  in  multi- 
plication and  division  can  be  given.  A  few  of  the  most  useful 
ones  are  given  here.  If  benefit  is  to  be  derived  from  them, 
they  must  be  very  carefully  fixed  in  mind,  and  used  whenever 
occasion  arises. 


54  PRACTICAL  MATHEMATICS 

(1)  To  multiply  a  number  by  5,  50,  500,  etc.,  multiply  by 
10,  100,  1000,  etc.,  and  divide  by  2. 

Why  will  this  give  the  result? 

Example.     7856  X  50  =  785600  -5-  2  =  392800.  A  ns. 

Multiply  tho  following  without  using  the  pencil: 
76X50  432X50  5.5X5 

96X5  768X500  4.35X50 

88X500  47X50  79.2X5000 

(2)  To  multiply  by  25,   250,   etc.,  multiply  by   100,   1000, 
etc.,  and  divide  by  4. 

Why  will  this  give  the  result? 

Example.     32  X  250  =  32000 -r  4  =  8000.  Ans. 

Multiply  the  following  without  using  the  pencil : 
256X25  8956X25  728X250 

74.92X250  492X2500  942.3X2500 

(3)  To  multiply  a  number  by   125,  multiply  by  1000  and 
divide  by  8. 

Why  will  this  give  the  result? 
Example.     848 X 1 25  =  848000 -f- 8  =  106000.  Ans. 
Multiply  the  following: 
920X125  4.76X125  72.88X125  55.5X125 

(4)  To  multiply  a  number  by  33$,  16f ,  12$,  8$,  or  &\,  multiply 
by  100  and  divide  by  3,  6,  8,  12,  or  16. 

Example.     84  X  8$  =  8400  -5-1 2  =  700.  Ans. 

Multiply  the  following: 

48X33*  42.6X16?  32*X16§  41|X8* 

96X12*  3.97X8*  33JX33*  19|X6J 

72X6J  4.76X33*  98.76X16?  27|X12J 

This  rule  can  be  used  easily  in  multiplying  a  number  by  37*,  62$, 
87*,  83},  and  other  fractional  parts  of  100  or  1000. 

Multiply  the  following: 

24X62*  35X333*  421x62* 

32X87*  476*X625  71$  X37* 

36X83*  672X62*  47«X333* 

64X37*  272X87*  36*  X83J 

(5)  To  multiply  a  number  ending  in  \,  as  2$,  4$,  11$,  by 
itself,   multiply  the  whole  number  by  the  whole  number  plus 
1  and  add  J  to  the  product. 


SHORT  METHODS  AND  CHECKS  55 


Examples.     8|X8£  =  8X9  +  |  =  72f.     Ans. 

ll^Xlii  =  11X12+|  =  132|.     Ans. 
The  reason  may  be  shown  as  follows: 


But  3X^+1X3=1X3  and  |X|  =  i 
Hence  3^X3^ 


Multiply  the  following: 


40JX40J  59|X59| 

Putting  in  the  decimal  form,  we  have 

81x8^  =  8.5X8.5  =  72.25. 

Now  removing  the  decimal  point,  we  have 

85X85  =  7225. 

Multiply  the  following  : 

7.5X7.5  135X135  505X505 

12.5X12.5  95X95  615X615 

11.5X11.5  155X155  925X925 

(6)  Divisions.  By  using  the  inverse  operations  to  those 
given  in  the  preceding  rules,  we  may  divide  by  33|,  16f, 
12|,  125,  250,  8i  etc. 

Make  the  rules  for  divisions. 

84  -M2|  =  84  -^100X8  =  6.  72. 


32-=-  125  =32-M  000X8  =  0.256. 
450  +  61i  =  450  -MOO  X  16  =  72. 
23  ^-  250  =  23  -M  000X4  =0.0092. 

The  multiplications  in  such  problems  can  usually  be  per- 
formed without  using  the  pencil. 

Divide  the  following  : 

800-M2*  492  -H  16f  720  -5-8-J- 

37.6  -f-  250  923  4-331  783  -M2i 

7.62  -=-12^  436^-3|  7.29  -=-125 

927^-333^  43.9-^250  8927  -^166f 

44.  Checking.  —  No  check  can  be  made  that  is  absolutely 
certain  to  detect  an  error,  but  there  are  many  very  useful 
devices  for  checking  the  accuracy  of  the  work. 

(1)  Addition.  A  simple  way  to  check  addition  is  to  re-add, 
taking  the  figures  in  some  other  order.  Add  first  up  and  then 
down,  is  very  satisfactory. 


56  PRACTICAL  MATHEMATICS 

(2)  Subtraction.     An  error  in  a  subtraction  will  generally 
be  detected   by  adding  the  remainder   to  the  subtrahend. 
If  this  gives  the  minuend  the  work  is  correct. 

Example.  Minuend  37249 
Subtrahend  18496 
Remainder  18753 

37249  =  subtrahend-^- remainder. 

(3)  Multiplication.     A  good  way  to  check  multiplication 
is  to  interchange  the  multiplicand  and  multiplier  and  multiply 
again. 

A  very  convenient  and  quick  method  is  to  proceed  a.s 
follows : 

(a)  Add  the  digits  in  the  multiplicand.  If  this  sum  has 
more  than  one  digit,  add  these.  Continue  till  a  number  of 
one  digit  is  found. 

(6)  Add  the  digits  of  the  multiplier  as  directed  in  (a). 

(c)  Multiply  together  the  numbers  obtained  in  (a)  and  (6), 
and  add  digits  till  a  number  of  one  digit  is  found. 

(d)  Add  digits  of  product  as  directed  in  (a). 

(e)  Compare  results  of  (c)  and  (d).     If  they  are  the  same 
the  work  checks. 

Check.  Example. 

Summing  digits  as  directed  34768  Multiplicand 

in  (a^  492  Multiplier 

for   multiplicand   gives   1,    for  69536 

multiplier    gives    6,    6X1=6.  312912 

Sum    of   digits   from    product  139072 

gives  6.     Since  this  is  the  same  17105856  Product 

as  obtained  before,  the  work  is 

checked. 

(4)  Division.     Division    can    be    checked    by    multiplying 
the  divisor  by  the  quotient  and  then  adding  the  remainder. 
The  result  should  be  the  dividend. 

A  quicker  way  to  check  is  to  add  the  digits  as  directed 
for  checking  multiplication:  (a)  the  dividend;  (6)  the  divisor; 
(c)  the  quotient;  (d)  the  remainder.  Multiply  the  results 
in  (6)  and  (c),  add  the  result  in  (d),  and  then  add  the  digits 
in  this  result  which  should  give  the  same  as  the  result  of  (a) 
if  the  work  is  correct. 


SHORT  METHODS  AND  CHECKS  57 

Check.  Example. 

Dividend  4923567476  Divisor 

(a)  Sum  of  digits,  dividend  9,  476        1 10343  Quotient 

(6)  sum  of  digits,  divisor  8,  1635 

(c)  sum  of  digits,  quotient  2, 

(d)  sum  of  digits,  remainder  2.  TO  !? 
8X2+2  =  18. 


1727 
Sum  of  digits  of  18  =  9,  which  ]428 

is  the    same    as    the  sum  in  "299  Remainder 

(a)  and  so  checks  the  work. 

The  preceding  rules  apply  as  well  to  decimals  as  to  whole 
numbers,  but  do  not  check  the  position  of  the  decimal  point. 

EXERCISES  14 

First  multiply  then  divide  the  following  and  check  as  directed  in  the 
preceding  article. 

1.  435678  by  4537.  6.  456.78  by  45.32. 

2.  980765  by  789.  7.  1230.8  by  3.876. 

3.  60385  by  4327.  8.  32418  by  8.098. 

4.  342153  by  7651.  9.  4.6543  by  1.0876. 
6.  45.654  by  345.  10.  32  654  by  7.547. 


CHAPTER  V 
WEIGHTS  AND  MEASURES 

45.  English  system. — The  English  system  of  weights  and 
measures  is  the  one  in  common  use  in  the  United  States.  The 
most  used  tables  and  equivalents  of  this  system  follow.  The 
problems  which  are  given  later  are  inserted  as  material  for 
review  of  work  which  it  is  supposed  the  student  has  done 
previously.  Suggestions  on  solutions  are  given  for  some  of  the 
exercises  that  follow  but  no  general  methods  are  given  as  to 
how  to  solve  such  exercises. 

(1)  Measures  of  time. 

60  seconds  (sec.)  =1  minute  (min.) 
60  minutes  =1  hour  (hr.) 

24  hours  =1  day  (da.) 

365  days  =  1  common  year  (yr.) 

366  days  =  1  leap  year. 

(2)  Measures  of  length. 

12  inches  (in.  or  ")  =  1  foot  (ft.  or  ') 

3  feet  =  1  yard  (yd.) 

5$  yards  =  1  rod  (rd.) 

320  rods  =  1  mile  (mi.) 

5280  feet  =  1  mile. 

1760  yards  =  1  mile. 

(3)  Measures  of  area. 

144  square  inches  (sq.  in.  or  in.2)  =  1  square  foot  (sq.  ft.  or  ft.z) 

9  square  feet  =  1  square  yard  (sq.  yd.  or  yd.1) 

30J  square  yards  =  1  square  rod  (sq.  rd.  or  rd.s) 

160  square  rods  =1  acre  (A.) 

640  acres  =1  square  mile  (sq.  mi.'* 

(4)  Measures  of  volume. 

1728  cubic  inches  (cu.  in.  or  in.3)  =  1  cubic  foot  (cu.  ft.  or  ft.1) 
27  cubic  feet  =  1  cubic  yard  (cu.  yd.  or  yd.1) 

128  cubic  feet  =  1  cord  (cd.) 

(5)  Liquid  measures. 

2  pints  (pt.)      =  1  quart  (qt.) 

4  quarts  =1  gallon  (gal.) 

31 J  gallons  =  1  barrel  (bbl.) 

231  cubic  inches  =  1  gallon. 
58 


WEIGHTS  AND  MEASURES  59 

(6)  Dry  measures. 

2  pints  (pt.)  =1  quart  (qt.) 

8  quarts  =1  peck  (pk.) 

4  pecks  =1  bushel  (bu.) 

2150.42  cubic  inches  =  1  bushel. 

It  should  be  carefully  noted  that  dry  and  liquid  measures  are  very 
different.  For  instance,  4  quarts  in  liquid  measure  contain  231  cu.  in., 
while  in  dry  measure  they  contain  268.8  cu.  in.  nearly. 

(7)  Measures  of  weight  (Avoirdupois). 

7000  grains  (gr.)  =  1  pound  (Ib.) 

16  ounces    (oz.)  =1  pound. 

100  pounds  =1  hundred  weight  (cwt.) 

2000  pounds  =1  ton  (T.) 

2240  pounds  =  1  long  ton. 

In  practice  it  is  customary  to  consider  1  cu.  ft.  of  water  as  62.5  Ib.  or 
1000  oz.  (See  Table  II.) 

EXERCISES  15 

1.  Reduce  27  yd.  2  ft.  11  in.  to  inches.  Ans.  1007  in. 

2.  Reduce  18  hr.  20  min.  35  sec.  to  seconds.  Ans.  66,035  sec. 

3.  Reduce  4  T.  7  cwt.  35  Ib.  9  oz.  to  ounces.  Ans.  139,769  oz. 

4.  Reduce  8  bu.  3  pk.  7  qt.  1  pt.  to  pints.  Ans.  575  pt. 
6.  Reduce  8  A.  25  sq.  rd.  4  sq.  yd.  to  square  yards. 

Ans.  39,480i  sq.  yd. 

6.  Reduce  5937  sq.  in.  to  higher  denominations. 

Ans.  4  yd.'-*  5  ft.2  33  in.2 

Suggestion.  First  divide  5937  by  144,  the  number  of  square  inches  in  a 
square  foot.  The  quotient  is  the  number  of  square  feet  and  the  re- 
mainder is  square  inches. 

Then  divide  by  the  number  of  square  feet  in  one  square  yard. 

7.  Multiply  12  cu.  yd.  15  cu.  ft.  1115  cu.  in.  by  6. 

Ans.  75yd.3  12ft.3  1506  in.3 

8.  How  many  iron  rails  each  30  ft.  long  will  be  required  to  lay  a  rail- 
road track  26  miles  long?  Ans.  9152. 

9.  Find  the  value  of  a  field  180  rods  long  and  94  £  rods  wide,  at  $18.00 
per  acre.  Ans.  $1913.625. 

10.  Reduce  17  pints  to  the  decimal  of  a  gallon.          Ans.  2.125  gal. 

11.  How  many  steps  does  a  man  take  in  walking  2  mi.  76  rd.  if  he 
goes  2  ft.  8}  in.  each  step?  Ans.  4362.1  - 

Suggestion.  Divide  the  total  number  of  inches  by  the  number  of  inches 
in  one  step. 

12.  Find  the  weight  of  1  gal.  of  water. 

Ans.  133.68+  oz.  =8  Ib.  5.68  oz. 

13.  How  many  sacks  each  containing  2  bu.  1  pk.  can  be  filled  from  a 
bin  containing  245  bu.?  Ans.   109  sacks  nearly. 


60  PRACTICAL  MATHEMATICS 

14.  A  large  steamship  will  hold  75  bargo  loads  of  wheat  at  8500  bu. 
to  the  bargo.  A  freight  car  40  ft.  long  will  carry  950  bu.  Find  the 
length  of  a  train  carrying  enough  wheat  to  load  the  steamship,  allowing 
2  ft.  between  cars. 

16.  What  decimal  part  of  a  foot  is  iV  in.  ?  j  in.  ?  What  decimal  part 
of  a  yard  is  each?  Ana.  0.0052 +  ;  0.03125;  0.00173 +  ;  0.010417 -. 

16.  Reduce  the  following  to  decimal  parts  of  a  foot:  (a)  1  in.,  (6)  2  in., 
(c)  3*  in.,  (d)  71  in. 

Ans.  (a)  0.0833+,  (6)  0.16G7-,  (c)  0.2917-,  (d)  0.61458+. 

17.  Reduce  each  in  exercise  16  to  a  decimal  part  of  a  yard. 

Ans.  (a)  0.02778 -,  (6)  0.0556-,  (c)  0.0972+,  (d)  0.20486+. 

18.  Reduce  the  following  to  decimal  parts  of  a  pound  avoirdupois: 
(a)  |  oz.,  (b)  li  oz.,  (c)  3  oz.,  (d)  7J  oz.,  (e)  13  oz.,  (/)  4J  oz. 

Ans.  (a)  0.046875,  (6)  0.09375,  (c)  0.1875,  (d)  0.46875, 

(e)  0.8125,  (/)  0.28125. 

19.  Reduce  3.36  in.  to  a  decimal  fraction  of  a  rod. 

Ans.  0.01697-  rd. 

20.  Reduce  a  pressure  of  22.5  Ib.  per  square  foot  to  ounces  per  squarr 
inch.  Ans.  2.5  oz. 

22  5x16 
Suggestion.     The  cancellation  is  — ~\TA, —  =2.5. 

21.  What  is  the  cost  per  hour  for  lighting  a  room  with  68  burners  each 
consuming  2J  cu.  in.  per  second,  the  price  of  gas  being  85  cents  per 
thousand  cubic  feet?  Ans.  27.1—  cents. 

22.  A  clock  that  gains  1  min.  in  10  hr.  is  correct  at  Monday  noon. 
What  is  the  correct  time  when  the  clock  registers  noon  on  the  following 
Monday?  Ans.  43  min.  14—  sec.  past  11  A.  M. 

23.  How  many  feet  per  second  are  equivalent  to  30  miles  per  hour? 

Ans.  44  ft. 

24.  If  sound  travels  at  the  rate  of  1125  ft.  per  second,  in  what  time 
would  the  report  of  a  gun  be  heard  when  fired  at  a  distance  of  1.276 
miles?  Ans.  5.989-  sec. 

26.  A  train  travels  316  miles  in  10  hr.  34  min.;  what  distance  will  it 
travel  in  27  hr.  17  min.  at  the  same  rate?  Ans.  815.9+  mi. 

26.  A  tank  holding  7  bbl.  has  2  pipes  opening  from  it;  one  empties  out 
2  qt.  in  5  sec.,  and  the  other  17  gal.  per  minute.     How  long  will  it  take  to 
empty  the  tank  if  both  pipes  are  open?  Ans.  9.587—  min. 

27.  If  it  takes  4  qt.  of  oats  for  one  feeding  for  a  horse;  how  many 
bushels  of  oats  will  it  take  to  feed  5  horses  one  year,  giving  them  2  feedings 
per  day?  Ans.  456}  bu. 

28.  A  carload  of  potatoes  has  a  total  weight  of  55,600  Ib.     The  car 
alone  weighs  15,675  Ib.     How  many  bushels  of  potatoes  in  the  carload 
if  potatoes  weigh  60  Ib.  per  bushel?  Ans.  665.41  bu. 

29.  A  tank  that  holds  25.6  bbl.  will  hold  how  many  bushels? 

Ans.  86.62+. 

30.  A  bin  that  holds  13  bu.  will  hold  how  many  gallons? 

Ana.   121.02-  gal. 


WEIGHTS  AND  MEASURES  61 

31.  How  long  would  it  take  a  cannon  ball  going  at  the  rate  of  1950  ft. 
per  second  to  reach  the  sun,  if  the  sun  is  distant  93,000,000  miles? 

Ans.  8  yr.  nearly. 

32.  Supposing  the  distance  travelled  by  the  earth  about  the  sun  to  be 
596,440,000  miles  per  year,  what  is  the  average  hourly  distance  travelled, 
taking  the  year  to  be  365J  days?     Find  the  average  distance  per  second. 

Ans.  19  miles  per  second,  nearly. 

33.  Find  the  area  in  acres  of  a  farm  which  is  represented  on  paper  as  a 
rectangle  3f  in.  by  10£  in.  on  a  scale  of  A  in.  to  the  rod. 

Ans.  63  acres. 

34.  If  4  oz.  of  the  white  of  egg  is  used  in  cleansing  50  gal.  of  wine, 
how  many  eggs  will  it  take  for  17  bbl.  of  wine?     One  egg  contains  1.1 
oz.  of  white.  Ans.  39  eggs. 

35.  The  total  cost  of  making  a  cement  walk  300  ft.  long,  5  ft.  wide,  and 
6  in.  thick,  where  the  cement  was  hand  mixed,  was  as  follows:  Foreman, 
8  hours;  laborers,  120  hours  at  a  cost  of  $53.20;  cement,  $86.00;  gravel, 
$34.08.     Find  the  total  cost  per  square  yard  and  per  square  foot. 

36.  A  farmer  drew  a  load  of  potatoes  to  market  for  which  he  received 
76  cents  a  bushel.     If  the  wagon  and  load  weighed  3710  Ib.  and  the 
empty  wagon  weighed  1150  Ib.,  find  what  he  received  for  the  potatoes. 
60  Ib.  of  potatoes  make  1  bu. 

37.  How  many  pounds  of  charcoal  does  it  take  to  make  3  tons  of  gun- 
powder, if  the  powder  is  rS  sulphur,  f  saltpeter,  and  the  rest  charcoal? 

Ans.  900. 

38.  How  many  barrels  of  flour,  196  Ib.  each,  does  it  take  to  run  a 
bakery  one   week  of  7  days  if  the  output  is  6000  loaves  a  day,  and 
there  are  9J  oz.  of  flour  in  each  loaf?  Ans.   127  bbl.  45|  Ib. 

39.  (a)  Find  the  number  of  cubic  feet  in  a  barrel  to  the  nearest  0.001. 
(6)  Find  the  number  of  cubic  feet  in  a  bushel  to  the  nearest  0.00001. 

Ans.  4.211  cu.  ft.;  1.24446  cu.  ft. 

40.  A  new  copper  cent  weighs  48  grains.     How  many  pounds  will  $50 
in  these  weigh?  Ans.  34 7  Ib. 

41.  One  of  the  largest  diamonds  in  the  world  weighs  3025f  carats. 
How  many  pounds  avoirdupois  is  this,  correct  to  the  nearest  0.0001? 
A  carat  is  3.168  grains.  Ans.  1.3694  Ib. 

42.  If  railroad  ties  are  placed  18  in.  apart  from  center  to  center,  how 
many  miles  will  54,320  ties  reach?  Ans.  1511  mi. 

43.  How  many  rails,  each  30  ft.  in  length,  are  used  in  laying  two 
railroad  tracks  from  New  York  to  Chicago  a  distance  of  870  mi.?     Find 
the  weight  of  these  rails  at  90  Ib.  per  yard. 

Ans.  612,480  rails;  275,616  tons. 

44.  Supposing  the  distance  from  the  earth  to  the  sun  to  be  91,713,000 
miles,  and  that  the  sun's  light  reaches  the  earth  in  8  min.  18  sec.,  what  is 
the  velocity  of  light  per  second?  Ans.  184,  Io3  miles  nearly. 

45.  The  pressure  of  the  atmosphere  is  14.7  Ib.  per  square  inch.     Find 
the  pressure  in  pounds  per  square  foot.  Ans.  2116.8  Ib. 


62  PRACTICAL  MATHEMATICS 

46.  A  column  of  water  how  high  will  give  a  pressure  of  1  Ib.  per  square 
inch?  Ana.  2.3  ft.  nearly. 

47.  If  the  ends  of  an  iron  beam,  bearing  5  tons  at  its  middle,  rest  upon 
stone  piers,  required  the  necessary  bearing  surface  of  each  pier  if  the  stone 
will  support  200  Ib.  per  square  inch  of  surface.  Ans.  25  in.1 

By  bearing  surface  is  meant  the  area  of  the  stone  in  contact  with 
the  beam. 

48.  One  voussoir  (or  block)  of  an  arch  ring  presses  its  neighbor  with  a 
force  of  50  tons.     If  the  joint  has  a  surface  of  5  sq.  ft.,  find  the  pressure 
per  square  inch.  Ans.  138.9—  Ib.  per  square  inch. 

49.  Work  is  done  when  resistance  is  overcome.     Ic  is  measured  by 
the  product  of  the  force  times  the  distance  over  which  the  force  acts. 
As  a  formula  this  is  w=fXs,  where  w  is  the  work,  /  the  force,  and  s  the 
distance.     If  the  force  is  in  pounds  and  the  distance  in  feet  then  the  work 
is  in  foot-pounds. 

A  steam  crane  lifts  a  block  of  granite  weighing  2  tons  80  ft.  Find  the 
work  done  in  foot-pounds.  Ans.  320,000  ft.  Ib. 

60.  How  many  foot-pounds  of  work  is  necessary  to  pump  100  bbl.  of 
water  to  a  height  of  120  ft.?     Use  1  bbl.  =4.211  cu.  ft. 

Ans.  3,158,203  ft.  Ib. 

61.  How  many  foot-pounds  of  work  is  done  in  lifting  an  elevator 
weighing  3  tons  to  the  top  of  a  building  220  ft.  high?     If  the  elevator  is 
raised  through  this  height  in  2  minutes,  how  many  foot-pounds  of  work  is 
done  per  second?     If  an  engine  of  one  horse-power  can  do  550  foot- 
pounds of  work  per  second,  an  engine  of  what  horse-power  will  be 
necessary  to  lift  the  elevator  to  the  top  in  2  minutes? 

62.  A  man  weighing  165  pounds  ascends  a  stairs  to  a  height  of  60  ft. 
in  20  seconds.     How  many  horse-power  does  he  exert? 

THE  METRIC  SYSTEM 

46.  From  a  study  of  weights  and  measures  in  the  United 
States,  it  is  seen  that  a  legal  standard,  the  troy  pound,  has 
been  established  for  the  use  of  the  mint;  but  that  beyond 
that,  our  weights  and  measures  in  ordinary  use  rest  on  custom 
only  with  indirect  legislative  recognition.  It  is  seen  that 
the  metric  weights  and  measures  are  made  legal  by  direct 
legislative  permission,  and  that  standards  of  both  systems 
have  been  equally  furnished  by  the  Government  to  the  several 
states;  that  the  customary  system  has  been  adopted  by  the 
Treasury  Department  for  use  in  the  custom-houses,  but  that 
the  same  department  has  by  formal  order  adopted  the  metric 
standards  as  "fundamental  standards"  from  which  measures 
of  the  customary  system  shall  be  derived. 


WEIGHTS  AND  MEASURES  63 

Commercial  intercourse  between  nations  makes  it  advisable,  if  not 
necessary,  to  have  a  uniform  system  of  weights  and  measures.  Such 
relations  cause  those  countries  not  already  using  the  metric  system  to 
make  more  and  more  use  of  that  system.  For  instance,  large  orders  for 
locomotives  placed  in  the  United  States  by  foreign  governments  or  by 
corporations  in  those  countries,  have  made  it  a  matter  of  good  business 
to  carry  out  the  manufacturing  in  metric  units  of  measure. 

The  European  war,  beginning  in  1914,  made  it  necessary  for  large 
manufacturing  plants  in  this  country  to  make  considerable  use  of  the 
metric  system.  It  is  stated  on  good  authority  that  the  first  six  months 
following  the  entrance  of  the  United  States  into  the  great  war  advanced 
the  use  of  the  metric  system  in  this  country  more  than  had  the  previous 
ten  years. 

A  few  manufacturing  companies  have  for  several  years  quoted  sizes 
of  reamers,  drills,  and  other  tools  in  the  metric  system  as  well  as  in  the 
ordinary  system.  At  present  this  custom  is  becoming  increasingly 
more  general. 

47.  Measure    of    length.     The    meter. — The    length    of 
the    meter    was    at    first    determined    as    one    ten-millionth 
part  of  the   distance   from   the  equator  to  the  north-pole. 
It     was    afterward    found    that    there    had    been    a    slight 
error  in  this   determination.     At   present  the  meter   is  the 
length    at  0°C.    of    a    certain    bar,    made    of    90    per    cent 
platinum  and  10  per  cent  iridium,  called  the  International 
Meter,  and  kept  at  the  International  Bureau  of  Weights  and 
Measures,  near  Paris. 

The  two  copies  of  the  meter  which  the  United  States  has 
are  made  of  the  same  material.  One  of  these  is  used  as  the 
working  standard,  and  the  other  is  kept  for  comparison. 
To  insure  still  greater  accuracy,  these  are  compared  at  regular 
intervals  with  the  International  Meter. 

48.  Legal  units. — As  has  been  stated  the  Treasury  Depart- 
ment has  determined  that  the  meter  shall  be  the  "  funda- 
mental  standard"    of   length.     By   the   act   of   July,    1866, 
Congress  fixed   the  relation,    1   meter  =  39.37   in.     This  is 
the   only  legal  relation   between   the   two   systems,    and   is 
used  in  the  Office  of  Standards  of  Weights  and  Measures  in 
this  country  in  deriving  the  inch,  foot,  yard,  etc.,  from  the 
meter.     Determined   in   this  way  the   customary  units   are 

gal. 
In  the  Philippine  Islands,  Porto  Rico,  and  Guam  the  metric 


64  PRACTICAL  MATHEMATICS 

system  is  in  general  use,  and  is  the  sole  legalized  system  for 
these  islands.1 

49.  Measure  of  surface. — There  is  no  fundamental  standard 
of  surfaces  or  areas  as  there  is  of  the  measures  of  length.  But 
as  the  measures  of  areas  are  based  upon  the  units  of  length, 
and  as  these  are  standards,  the  measures  of  areas  may  be  so 
considered. 

60.  Measures  of  volume.     Cubic  and  capacity  measures.— 
In  the  United  States  the  fundamental  standards  of  volume  are: 
(1)  the  cubes  of  the  linear  units  based  on  the  International 
Meter;  (2)  the  liter,  which  is  the  volume  of  the  mass  of  one 
kilogram  of  pure  water  at  its  greatest  density;  (3)  the  gallon, 
which  is  231  cu.  in.;  (4)  the  bushel,  which  is  2150.42  cu.  in. 
The  liter  here  used  is  almost  exactly  1  cubic  decimeter,  and  the 
inch  is  derived  from  the  meter  according  to  the  relation, 
1  meter  =  39.37  in. 

61.  Measures   of   mass. — The   fundamental   standard   of 
mass  (weight)  in  the  United  States  is  the  International  Kilo- 
gram, a  cylinder  of  90  per  cent  platinum  and  10  per  cent 
iridium,  preserved  at  the  International  Bureau  of  Weights  and 
Measures,  near  Paris.     As  in  the  case  of  the  meter,  one  of  the 
two  copies  of  the  kilogram  possessed  by  the  United  States  is 
used  as  a  working  standard,  and  the  other  is  kept  under  seal 
and  used  only  to  compare  with  the  working  standard  from 
time  to  time.     To  insure  still  greater  accuracy,   these  are 
compared  at  regular  intervals  with  the  International  Kilogram. 

By  act  of  Congress  of  July  28,  1866,  the  pound  is  derived 
from  the  kilogram.  The  relation  established  at  that  time  was 
1  kilogram  =  2.2046  avoirdupois  pounds.  This  relation  has 
since  been  made  more  nearly  accurate  and  is  1  kilogram  = 
15,432.35639  grains,  which  would  change  the  first  relation  to  1 
kilogram  =  2.20462234  Ib.  avoirdupois,  or  1  Ib.  avoirdupois  = 
453.5924277  grams.  This  value  is  the  one  used  by  the 
National  Bureau  of  Standards  in  Washington.  It  is  thus 
seen  that  the  avoirdupois  pounds,  ounces,  etc.,  in  common  use 

1  See  Introduction  to  "Laws  Concerning  Weights  and  Measures  of  the 
United  States,"  compiled  by  Louis  A.  Fisher  and  Henry  D.  Huhbard  of 
the  Bureau  of  Standards,  Washington,  D.  C. 


WEIGHTS  AND  MEASURES  65 

are  derived  from  the  kilogram,  and  so  are  fixed  and  definite 
derived  units. 

The  established  relation  between  the  troy  pound  and  the 
avoirdupois  pound  is  1  troy  pound  =  fM$  avoirdupois  pounds. 

When  made,  the  standard  kilogram  was  supposed  to  be  the 
exact  mass  of  one  cubic  decimeter  or  1  liter  of  pure  water  at 
the  temperature  of  its  greatest  density.  It  has  been  found 
that  this  is  not  exactly  true,  but  the  difference  is  very  slight, 
the  kilogram  being  about  25  parts  in  1,000,000  too  heavy. 
This  difference  is  so  very  small  that  it  could  hardly  affect  any 
ordinary  problem. 

52.  Tables  and  terms  used. — In  the  customary  system  of 
weights  and  measures  we  have  about  150  different  terms  and 
50  different  numbers,  ranging  all  the  way  from  2  to  1728. 
These  numbers  bear  no  relation  to  one  another.  In  the 
metric  system  we  have  only  14  different  terms  and  but  a 
single  number,  and  that  is  the  number  10. 

In  the  metric  system  the  different  terms  used  are  the 
following : 

meter — the  unit  of  length, 
liter — the  unit  of  volume, 
are — the  unit  of  area, 
gram — the  unit  of  weight, 
myria — which  denotes  10,000, 
kilo — which  denotes  1000, 
hecto — which  denotes  100, 
deka — which  denotes  10, 
deci — which  denotes  0.1, 
centi — which  denotes  0.01, 
milli — which  denotes  0.001. 

Terms  which  are  sometimes  used  are: 
millier— which  denotes  1,000,000, 
quintal — which  denotes  100,000, 
stere — which  is  1  cubic  meter. 

To  these  might  be  added  mikron  and  mikrogram. 
If  the  foregoing  terms  are  carefully  fixed  in  mind  the  tables 
are  easily  formed. 


t)6 


PRACTICAL  MATHEMATICS 


(1)  Measures  of  length. 
10  millimeters  (mm.)   =  1  centimeter  (cm.) 
10  centimeters 
10  decimeters 
10  meters 
10  dekametei-s 
10  hectometers 


=  1  decimeter  (dm.) 
=  1  meter  (.m.) 
=  1  dekameter  (Dm.) 
=  1  hectometer  (.Hm.) 
=  1  kilometer  (Km.) 


=  0.01  meter 
=  0.1  meter 

>  10  meters 
••  100  meters 
=  1000  meters 


10  kilometers 


=  1  myriameter  (Mm.)     =10,000  meters 


(2)  Measures  of  surface. 

100  square  millimeters  (mm.2)  =1  sq.  centimeter  (cm.1) 

100  square  centimeters  =1  sq.  decimeter  (dm.2) 

100  square  decimeters  =1  sq.  meter  (m.2)  =  l  centare  (ca.) 

100  square  meters  =1  sq.  dekameter  (Dm.2)  =  1  are  (a.) 

100  square  dekameters  =  1  sq.  hectometer  (Hm.*)  =  1 

hektare  (Ha.) 

100  square  hectometers  =1  sq.  kilometer  (Km.*) 

(3)  Measures  for  land. 

100  centares  (ca.)  =1  are  (a.) 

100  ares  =  1  hectare  (Ha.) 

(4)  Measures  of  volume. 

1000  cubic  millimeters  (mm.3)  =  1  cu.  centimeter  (cm.3  or  cc.) 

1000  cubic  centimeters  =1  cu.  decimeter  (dm.1)  =  1  liter  (1.) 

1000  cubic  decimeters  =  1  cu.  meter  (m.3)  =  1  kiloliter  (Kl.) 


(5)  Measures  of  capacity. 
10  milliliters  (ml.)        =  1  centiliter  (cl.) 
10  centiliters 
10  deciliters 
10  liters 
10  dekaliters 
10  hectoliters 


=  1  deciliter  (dl.) 

=  1  liter  (!.)  =  !  dm.* 

=  1  dekaliter  (Dl.) 

=  1  hectoliter  (HI.) 

=  1  kiloliter  (Kl.)  =  lm.» 


(6)  Measures  of  weight. 
10  milligrams  (mg.)      =  1  centigram  (eg.) 
10  centigrams 
10  decigrams 
10  grams 
10  dekagrams 
10  hectograms 
10  kilograms 
10  myriagrams 
10  quintals 


=  1  decigram  (dg.) 

=  1  gram  (g.) 

=  1  dekagram  (Dg.) 

=  1  hectogram  (Hg.) 

=  1  kilogram  or  kilo  (Kg.) 

=  1  myriagram  (Mg.) 

=  1  quintal  (Q.) 

=  1  millier,  tonneau,  or  metric  ton  (T.) 


To  these  may  be  added  the  following  used  in  scientific  work : 
1  mikron  (M)  =0.000001  meter. 

1  mikrogram  (7)  =0.000001  gram. 


WEIGHTS  AND  MEASURES 


67 


Note.  A  chart  showing  very  clearly  the  relations  of  the 
different  measures  can  be  secured  by  addressing  the  Bureau 
of  Standards,  Washington,  D.  C. 


1 

i   i 

'i 

11          '     '     '     '2 

I 

1 

'3 

1     1 

1 

Inches 

Centimeters 

Illl 

i 

u 

2 

Illl 

345 

illl 

6 

M 

7 
llllll 

Illll 

v> 

ill 

Illl 

9 

FIG.  14. 

63.  Equivalents. — To  be  remembered. 

1  gallon  =231  cubic  inches  (established  by  law). 

1  bushel  =2150.42  cubic  inches  (established  by  law). 

1  meter  =39.37  inches  (established  by  law). 

1  gram  =  15.432  grains. 

1  pound  avoirdupois  =7000  grains. 

1  inch  =2.54  centimeters  (approximately). 

1  kilogram  =2.2  pounds  (approximately). 


For  Reference. 

Lengths 


1  inch 
1  foot 

1  kilometer 
1  mile 

1  sq.  in. 
1  sq.  ft. 
1  sq.  yd. 
1  cm.2 
1  m.2 
1  are 
1  acre 


Areas 


=  2.54001  cm. 

=  30.4801  cm. 

=  3280.83  ft,  =0.62137  mi. 

=  1.60935  Km. 

=  6.45163  cm.2 

=  0.0929034m.2 

=  0.836131  m.2 

=  0.155  sq.  in. 

=  10.76387-  sq.  ft.  =1.19599-  sq.  yd. 

=  119. 5985  sq.  yd. 

=  40.4687  ares. 


Volumes,  capacities 


1  cu.  in. 
1  cu.  ft. 
1  pt,  (liquid) 
1  pt.  (dry) 
1  qt.  (liquid) 
1  qt.  (dry) 
1  cm.3 
1  liter 
1  liter 
1  liter 


=  16.38716  cc. 

=  28.317  liters  or  dm.3 

=  473.179  cc.  =0.473179  liters  or  dm.3 

=  550.614  cc.  =0.550614  liters  or  dm.3 

=  946.358  cc.  =0.946358  liters  or  dm.3 

=  1101.228  cc.  =1.101228  liters  or  dm.3 

=  0.0610234  cu.  in. 

=  61.0234  cu.  in. 

=  2.11336  pt.  (liquid)  =  1.81616  pt.  (dry). 

=  1.05668  qt.  (liquid)  =0.90808  qt.  (dry). 


f>8  PRACTICAL  MATHEMATICS 

Weights  (maw) 

1  grain  -0.0647989  gram. 

1  ounce  (avoirdupois)  =28.3495  grams. 

1  pound  (avoirdupois)  =453.5924277  grams  - 0.45359  +  Kg. 

1  ton  (short)  =907.185  kilograms. 

1  gram  =  15.43235639  grains. 

1  kilogram  =2.20462  pounds  (avoirdupois). 

1  metric  ton  =2204.62  pounds  (avoirdupois). 

64.  Simplicity  of  the  metric  system. — The  men  who  devised 
the  metric  system  endeavored  to  invent  a  system  of  weights 
and  measures  that  would  be  as  simple  as  possible;  and  they 
undoubtedly  succeeded  in  making  a  system  that  is  simpler 
than  any  other  in  use. 

Many  look  upon  the  system  as  difficult  because  they  con- 
sider the  difficulties  of  changing  from  the  English  to  the  metric 
system,  or  from  the  metric  to  the  English,  as  difficulties  of 
the  metric  system.  In  reality  this  is  not  the  case,  as  all  such 
difficulties  would  disappear  if  the  metric  system  were  in  univer- 
sal use. 

The  simplicity  of  the  metric  system  lies  in  the  two  facts: 
first,  it  is  decimal,  and  therefore  fits  our  decimal  notation; 
second,  its  units  for  lengths,  surfaces,  solids,  and  weights 
are  all  dependent  on  one  unit,  the  meter. 

Ability  to  handle  the  metric  system  easily,  depends,  in 
great  part,  on  understanding  thoroughly  the  terms  used. 
It  is  of  first  importance  then  to  learn  well  these  terms  and  their 
meanings.  For  instance,  the  word  decimeter  should  mean, 
at  once,  one-tenth  of  a  meter. 

Because  of  the  decimal  relations  between  the  different 
terms  used,  the  changing  from  one  unit  to  another  is  a  very 
simple  matter.  In  reducing  to  higher  denominations,  we 
divide  by  10,  100,  1000,  etc.,  by  removing  the  decimal  point 
to  the  left. 

Thus,  to  change  3768  cm.  to  meters,  we  divide  by  100  by  removing 
the  decimal  point  two  places  to  the  left,  and  have  3768  cm.  =37.68  m. 
In  a  similar  manner,  72,468  g.  =72.468  Kg.,  and 
8643  1-86.43  HI. 

It  should  be  noticed  that  we  never  write  4  Km.  7  Hrn. 
3  Dm.  5  m.  but  write  it  4735  m.  The  former  way  of  writing 


WEIGHTS  AND  MEASURES  69 

it  would  be  similar  to  writing  $7.265  in  the  form  7  dollars  2 
dimes  6  cents  5  mills. 

In  reducing  to  lower  denominations,  the  multiplication 
is  performed  by  moving  the  decimal  point  to  the  right. 

Thus,  25  m.  =250  dm.  =25,000  mm.  and  16  Kg.  =16,000  g. 

55.  Relations  of  the  units. — It  cannot  be  impressed  upon 
the  mind  of  the  student  too  strongly  that  he  should  understand 
clearly  the  relations  between  the  units  of  different  kinds  of 
measure.     He  must  know  that  a  liter  is  a  cubic  decimeter, 
that  a  kilogram  is  the  weight  of  a  liter  of  pure  water,  that  an 
are  is  a  square  dekameter,  and  so  on.     He  should  notice  that 
in  the  surface  measures,  when  using  square  meters,  dekameters, 
etc.,  the  scale  is  100;  while  in  using  cubic  meters,  dekameters, 
etc.,  for  volumes,  the  scale  is  1000. 

Thus,    2m.2  =  200    dm.2  =  20,000  cm.2 
and    3m.3  =3000  dm.3  =3,000,000  cc.  =3,000,000,000  mm.3 

56.  Changing  from  English  to  metric  or  from  metric  to 

English  systems. — The  changing  from  one  system  to  another 
is  simply  a  matter  of  multiplication  or  division. 

(1)  Thus,  to    express  17  m.  in  inches, 

1m.  =39.37  in. 

17  m.  =  39.37  in.  X 17  =  669.29  in. 

(2)  Also,  to  express  2468  Ib.  in  kilograms, 

2.2  Ib.  (approx.)  =  1  Kg. 
2468  Ib.  =  2468  -5-  2.2  =  1 121 .8  Kg. 
Or  using  the  equivalent  1  Ib.  =  0.45359  Kg., 

2468  Ib.  =  0.45359  Kg.  X 2468  -  1 1 19.46  Kg. 
The  disagreement  in  the  results  is  on  account  of  2.2  Ib.  being 
a  rough  approximation. 

The  United  States  Bureau  of  Standards  has  compiled 
numerous  tables  of  equivalents  for  use  in  the  custom  houses. 
By  the  use  of  these  tables,  a  conversion  from  one  system  to 
another,  is  made  by  simply  referring  to  the  proper  table  and 
reading  the  result. 

For  further  information  the  following  pamphlets  can  be 
obtained  gratis  from  the  Bureau  of  Standards,  Washington, 
D.  C.:  History  of  Standard  Weights  and  Measures  of  the 


70  PRACTICAL  MATHEMATICS 

United  States,  Table  of  Equivalents,  and  the  International 
Metric  System  of  Weights  and  Measures. 

EXERCISES  16 

1.  Express  the  following,  first,  in  meters,  and  second,  in  millimeters: 
456  cm.,  1763  Dm.,  27  Km.         Ans.  4.56  m.,  17,630  m.,  27,000  m., 

4560  mm.,  17,630,000  mm.,  27,000,000  mm. 

2.  Express  the  following  in   m.2:  75  cm.1,    125  mm.1,   0.025   Dm.*, 
0.0029  Km.s  Ans.  0.0075  m.»,  0.000125  m.1,  2.5  m.1,  2900  m.» 

3.  Expres    the  following  in  terms  of  m.3:  1756  1.,  467  KL,  4937  dl., 
0.1067    Dl.,    735,432    dm.',    764    Dm.»    Ans.  1.756    m.»,    467    m.1, 

0.4937  m.3,  10.67  m.3,  735.432  m.3,  764,000  m.» 

4.  Reduce  750  1.  to  liquid  quarts;  326  1.  to  dry  quarts;  75  m.  to  inches; 
576  cm.  to  feet;  27  m.3  to  bushels;  9276  mm.3  to  gallons;  12  Dm.3  to 
barrels.         Ans.  792.51   qt.,  296.03408  qt.,  2952.75  in.,   18.8976  ft, 

766.19  bu.,  0.00245  gal.,  100,636.19  bbl. 
Solution.     From  Art.  47  find  1  1.  =1.05668  qt.  (liquid). 

.'.  7501.  =1.05668  qt.X 750  =  792. 51  qt.  Ans. 

In  the  fifth  part,  27  m.3  to  bushels,  the  change  is  not  so  easy  from  the 
equivalents  given. 

27  m.3  =27,000  dm.3  or  liters. 

11.  =0.90808  qt.  (dry). 
/.  27,000  1.  =0.90808  qt.  X27,000  =  24,518. 16  qt. 

Divide  this  by  32  because  1  bu.  =32  qt. 
.'.  27  m.3  =  766.19  bu.  Ans. 

6.  Reduce  456  in.  to  meters;  43.5  ft.  to  centimeters;  327  gal.  to  liters; 
92.87  qt.  (dry)  to  liters;  756  bu.  to  cubic  meters; 

Ans.   11.58  m.,  1325.88  cm.,  1237.84  1.,  102.27  1.,  26.64  m.3 

6.  No.  16  gage  sheet  steel  is  -?6  in.  thick  and  weighs  40  oz.  per  square 
foot.     Find  thickness  in  millimeters  (four  decimal  places),  and  the  weight 
per  square  meter  in  kilograms  (two  decimal  places). 

Ans.  1.5875  mm.,  12.21  -Kg.  per  m.2 

Solution.     To  find  the  weight  in  Kg.  per  m.2,  first  find  the  weight  of  a 
square  meter  in  ounces  and  then  change  to  pounds  and  to  kilograms. 

1  m.2  =  10.76387ft.2 

.'.  1  m.2  weighs  40  oz.  X 10. 76387  =430.5548  oz.  =26.9097  Ib. 
1  Ib.  =0.45359  Kg. 
.-.26.9097  Ib.  =0.45359  Kg.  X26.9097  =  12.20597  Kg.  Ans. 

7.  No.  24  gage  sheet  steel  is  0.635  mm.  thick  and  weighs  4.882  Kg. 
per  m.2     Find  thickness  in  decimal  of  inch  (three  decimal  places),  and 
weight  per  ft.2  in  ounces.  Ans.  0.025  in.,  16  oz.  per  ft.2 

8.  Find  the  difference  between  313  in.  and  10  cm. 

Ans.  3U  in.  larger  by  0.0005  in. 

9.  Feb.   12,   1912,  Oscar  Mathieson,  of  Norway,  set  a  new  world's 
record  in  ice  skating.     He  made  1500  m.  in  2  min.  20  sec.     This  is  a 
mile  in  what  time? 


WEIGHTS  AND  MEASURES  71 


Solution.     2  min.  20  sec.  =  140  sec. 

1500X39.37,^ 

1500  m.  = —j ft. 

\2t 

1500X39.37 

—  0         n —  =  number  of  feet  m  1  sec. 
1^  X 14U 

5280X12X140 
1500X39.37 

150.2  sec.  =2  min.  30.2  sec.  Ans. 


=  150.2  =  number  of  sec.  to  go  1  mi. 


10.  The  same  day  in  Chicago,  Harry  Kaad  won  the  mile  race  in  3  min. 
23!  sec.     This  is  1500  m.  in  what  time?  Ans.  3  min.  9.6  sec. 

11.  In  describing  the  making  of  reenforced  concrete  the  necessary 
pressure  is  given  as  25  kilos  per  square  centimeter.     How  many  pounds 
is  this  per  square  inch?  Ans.  355.58+. 

12.  Find  in  kilograms  the  weight  of  air  in  a  room  10.5  by  8.3  by  4 
meters,  air  being  0.001276  times  as  heavy  as  water. 

Ans.  444.8136  Kg. 

13.  Find  the  weight  in  kilograms  of  the  mercury  in  a  tube  of  1  cm.2 
cross  section  and  760  mm.  long,  mercury  being  13.596  times  as  heavy  as 
water.  Ans.  1.0333-  Kg. 

14.  If  a  map  is  made  on  a  scale  of  1  to  60,000,  how  many  kilometers 
do  79  mm.  on  the  map  represent?  Ans.  4.74. 

15.  If  a  person  in  breathing  uses  0.25  m.3  of  air  a  minute,  how  long 
will  it  take  6  persons  to  use  the  air  in  a  room  6  m.  long,  3.5  m.  high,  and 
5.3  m.  wide?  Ans.  74.2  min. 

16.  A  block  of  stone  weighs  7643  Kg.     A  cubic  decimeter  of  the  stone 
weighs  2.7  Kg.     Find  the  volume  of  the  block  in  cubic  meters. 

Ans.  2.83074  m.3 

17.  Find  the  area  in  hectares  of  a  triangular  field  whose  base  is  70  m. 
and  the  altitude  60  m.  Ans.  0.21  Ha. 

18.  How  many  liters  of  water  are  contained  in  a  reservoir  10  m.  X6 
m.  X4  m.?     What  is  the  weight  of  the  water  in  kilograms? 

Ana.  240,000  1.,  240,000  Kg. 

19.  Find  the  capacity  in  liters  of  a  rectangular  tank  2  m.  X9  dm.  X8 
dm.  Ans.   1440  1. 

20.  What  is  the  length  of  a  centigram  of  wire  255  mm.  of  which  weighs 
0.172g.?  Ans.  14.83-  mm. 

21.  A  liter  of  mercury  weighs  13.596  Kg.;  how  many  mm.3  of  mercury 
weigh  1  g.?  Ans.  73.551. 

22.  A  man's  height  is  174  cm.     What  is  his  height  in  feet  and  inches? 

Ans.  5  ft.  8.5+  in. 

23.  Express  the  following  readings  in  centimeters:  29.9  in.,  30.0  in., 
30.1  in.,  30.2  in. 

Ans.  75.946  cm.,  76.200  cm.,  76.454  cm.,  76.708  cm. 

24.  Express  the  following  in  inches  *o  the  nearest  0.01:  71.119  cm., 
73.659  cm.,  74.929  cm.  Ans.  28.00  in.,  29.00  in.,  29.50  in. 


72  PRACTICAL  MATHEMATICS 

26.  Cast  copper  being  8.8  times  as  heavy  aa  an  equal  volume  of  water. 
what  is  the  weight  of  5  cm.J?  An*.  44  g. 

26.  A  velocity  of  32.2  ft.  per  second  is  how  many  centimeters  per 
second?  An*.  981.5 -. 

27.  A  rate  of  1  mile  in  2  min.  6  sec.  is  how  many  kilometers  per  minute? 
How  many  meters  per  second?  Ans.  0.7664—,  12.77+. 

28.  A  rate  of  30  miles  per  hour  is  at  the  rate  of  one  kilometer  in  how 
many  minutes?  An*.  1.243  —  . 

29.  A  pressure  of  14.7  Ib.  per  square  inch  is  how  many  grams  per  cm.1? 

Solution.  7=  =  number  of  Ib.  per  sq.  cm. 

O.451DO 

14.7X453.5924 

.-,.._     -  =  1033.5+  =  number  of  g.  per  sq.  cm. 
D.  451  Go 


CHAPTER  VI 
PERCENTAGE  AND  APPLICATIONS 

57.  Per    cents    as    fractions. — To    one    who    thoroughly 
understands  fractions,  percentage  offers  no  new  difficulties. 

The  words  per  cent  mean  by  the  hundred.  The  symbol 
%  means  per  cent.  Thus,  10%  means  10  per  cent  or  iVV 
or  0.10.  In  a  similar  way: 

5%  =0.05  =.}a  50%  =0.50  =\ 

10%  =0.1  =1I0  60%  =0.60  -$ 

12J%  =0.12*  =|  62i%  =  0.62^  =f 

16|%  =0.16§  =  £  75%  =0.75  =f 

20%  =0.20  =  !,-  80%  =0.80 

25%  =0.25  =1  83^%  =0.83^  =  1} 

33^%  =0.33^  =J  87i%  =0.87|  =| 

37i%  =0.37*  =|  90%  =0.90  =A 

40%  =0.40  =| 

To  change  a  fraction  as  |  to  an  equivalent  form  in  per 
cent,  reduce  it  to  a  fraction  having- 100  for  a  denominator. 

Thus,  1  =  TY<r  =  40%; 
or  I    =  |  of  100%  =  40%. 
Similarly  I  =  f  of  100%  =  87$%. 

A  per  cent  expressed  as  |%,  does  not  mean  |  but  |  of 
1%,  which  is  the  same  as  f  of  1^5-  =  ?|7  =  -j^  =  0.004. 
In  the  same  way  f  %  =  f  of  y|~¥  =  ¥-|T  =  0.00375  = 
0.375%. 

It  should  be  carefully  noticed  that  the  sign  %  does  the 
duty  of  two  decimal  places. 

Thus,  0.05  =  5%,  0.0005  =  0.05%,  1.07  =  107%,  and  4.33$ 
=  433$%. 

58.  Cases. — The  problems  of  percentage  usually  occurring 
are  of  the  following  forms: 

(1)  What  is  37£%  of  720? 

(2)  45  is  what  per  cent  of  450? 

(3)  85  is  62$%  of  what  number? 

73 


74  PRACTICAL  MATHEMATICS 

These  three  forms  can  be  stated  in  general  terms  if  the 
following  definitions  are  given : 

The  number  of  which  the  per  cent  is  taken  is  the  base. 

The  number  of  per  cent  taken  is  called  the  rate. 

The  part  of  the  base  determined  by  the  rate  is  the  per- 
centage. 

The  sum  of  the  base  and  percentage  is  the  amount. 

The  base  minus  the  percentage  is  the  difference. 

The  three  problems  now  become  the  cases: 

Case  I.  Base  and  rate  given  to  find  percentage. 

Case  II.  Base  and  percentage  given  to  find  rate. 

Case  III.  Percentage  and  rate  given  to  find  base. 

These  three  cases  of  percentage  correspond  to  the  throe 
cases  in  multiplication;  when  any  two  of  the  numbers,  multi- 
plicand, multiplier,  and  product,  are  given,  to  find  the  third. 

multiplicand  corresponds 
to  base 


In  multiplication, 


multiplier  corresponds  to 


>  in  percentage. 


rate 
product    corresponds    to 

percentage 

Case  I  corresponds  to:  multiplicand  and  multiplier  given 
to  find  the  product. 

Product  =  multiplicand  X  multiplier. 

Case  II  corresponds  to:  multiplicand  and  product  given  to 
find  the  multiplier. 

Multiplier  =  product  -f-  multiplicand. 

Case  III  corresponds  to:  multiplier  and  product  given  to 
find  the  multiplicand. 

Multiplicand  =  product  -f-  multiplier. 

69.  Rules  and  formulas. — In  the  language  of  percentage 
these  become: 

Case  I.     Percentage  =  base  X  rate. 

This  may  be  written  as  a  formula  if  b  stands  for  base,  p 
for  percentage,  and  r  for  rate.     The  formula  is 

p  =  b  X  r. 
Case    II.     Rate  =  percentage  -=-  base.     The  formula  is 

r  =  p  -f-  6. 

Case    III.     Base  =  percentage  -r-  rate.     The    formula    is 
6  =  p  -*•  r. 


PERCENTAGE  AND  APPLICATIONS  75 

60.  Solutions.     Problem  (1)  is  solved  thus: 

By  fractions.     37|%  of  720  =  f  of  720  =  270.  Ans. 

By  formula.  Using  the  formula  p=  bX  r,  gives  the 
same  result,  for  then  p  =  720  X  0.37£  =  270.  Ans. 

Problem  (2).  By  fractions.  45  is  what  per  cent  of  450 
means  45  is  how  many  hundredths  of  450,  that  is,  some 
number  of  hundredths  of  450  is  45. 

Then  45  is  ^  =  yV  =  iVir  =  10%  of  450. 

By  formula,     r  =  p  -4-  6  gives  r  =  45  -f-  450  =  0.1  =  10%. 

Problem  (3).  By  fractions.  85  is  62^%  of  what  number 
is  the  same  as  85  is  f  of  what  number.  It  is  now  a  simple 
problem  in  fractions  and  may  be  reasoned  thus : 

If  85  is  f  of  some  number  then  17  is  f  of  that  number, 
and  136  is  f  of  that  number. 

Hence  the  number  =  136.  Ans. 

By  formula,     b  =  p  -f-  r  gives  p  =  85  -H  0.62^  =  136.  Ans. 

EXERCISES  17 

Solve  the  following  exercises  without  a  pencil  if  possible. 

1.  What  is  J  of  24?     0.25  of  24?     25%  of  24? 

2.  What  is  f  of  36?     0.75  of  36?     75%  of  36? 

3.  What  is  $  of  45?     0.33$  of  45?     33$%  of  45? 

4.  What  is  f  of  48?     0.66f  of  48?     66f%  of  48? 
6.  What  is  l  of  60?     0.20  of  60?     20%  of  60? 

6.  What  is  f  of  70?     0.80  of  70?     80%  of  70? 

7.  What  is  \  of  72?     0.12*  of  72?     12$  %  of  72? 

8.  What  is  f  of  48?     0.625  of  48?     62.5%  of  48? 

9.  What  is  ^  of  64?     0.06i  of  64?     6i%  of  64? 

10.  What  is  25%  of  16?     of  48?     of  90?     of  240? 

11.  What  is  33$%  of  75?     of  42?     of  96?     of  720? 

12.  What  is  4%  of  25?    of  75?    of  300?     of  1000? 

13.  What  is  7%  of  20?     of  14?     of  55?     of  300? 

14.  5  is  what  per  cent  of  10?     of  20?      of  40? 

15.  8  is  what  per  cent  of  16?     of  40?     of  80? 

16.  30  is  what  per  cent  of  90?     of  240?     of  360? 

17.  What  %  is  8  of  150?     7$  of  12?  Ans.  5$%;  62$%. 

18.  What  %  is  f  of  12$?     27i  of  600?  Ans.  4*%;  4|%. 

19.  20%  of  what  number  is  3?     7?     14?     17? 

20.  33$%  of  what  number  is  7?     8?     14?     90? 

21.  62$%  of  what  number  is  5?     20?     f?     H? 

22.  37$%  of  a  number  is  72;  find  the  number.  Any.   192. 

23.  20%  off  of  what  number  leaves  48?  Ans.     60. 
Suggestion.     20%  off  a  number  leaves  80%  of  the  number. 


76  PRACTICAL  MATHEMATICS 

24.  30%  off  of  what  number  leaves  60?  Ant.  71$. 

25.  33  J%  off  of  what  number  leaves  12?  Ant.     18. 

26.  68  is  15%  less  than  what  number?  Ans.     80. 

27.  49  is  30%  less  than  what  number?  Ans.     70. 
2Q.  18  is  80%  more  than  what  number?  Ans.     10. 

29.  80  is  33  j%  more  than  what  number?  Ans.     60. 

30.  98  is  40%  more  than  what  number?  Ans.     70. 

31.  87  J  is  37i%  less  than  what  number?  Ans.  140. 

32.  If  oranges  that  cost  25  cents  a  dozen  are  sold  at  3  for  10  cents 
what  part  of  the  cost  is  gained?     What  per  cent? 

33.  Pencils  are  bought  at  15  cents  a  dozen  and  sold  foi  2  cents  each. 
What  part  of  the  cost  is  gained?     What  per  cent?  Ant.  60%. 

34.  Bought  a  horse  for  $75  and  sold  it  for  $100;  find  the  gain  per  cent. 

Ans.  33$%. 

35.  Find  the  gain  per  cent  in  each  case  if  a  horse  was  bought  at  the 
following  prices:  $50,  $40,  $25,  $10,  $5,  and  $1 ;  and  sold  for  $100.     Find 
the  gain  per  cent  if  the  horse  was  given  to  the  seller. 

36.  I  bought  a  bicycle  for  $100  and  after  using  it  one  year  sold  it  for 
$55.     Find  the  per  cent  of  discount.  Ans.  45%. 

37.  A  gas  bill  was  25%  higher  last  month  than  this.     If  it  is  $6.40 
this  month,  what  was  it  last  month? 

38.  A  horse  was  bought  for  $100  and  sold  for  $90.     What  was  the  loss 
per  cent? 

39.  A  man  sold  a  suit  of  clothes  gaining  i  the  cost.     What  part  of  the 
cost  was  the  selling  price?     What  was  the  gain  per  cent?     What  per 
cent  of  the  selling  price  was  the  cost? 

40.  A  quantity  of  wool  was  bought  for  $360,  and  f  of  it  was  then  sold 
for  the  cost  of  the  whole.     \Vhat  per  cent  would  have  been  gained  if  tho 
entire  amount  had  been  sold  at  the  same  rate?  Ans.  33J%. 

41.  A  man  spent  16§%  of  his  salary  for  board  and  room.     If  he  spent 
$6.50  a  week  for  board  and  room,  what  was  his  yearly  salary?     (52 
weeks  per  year.)  Ans.  $2028. 

61.  Applications. — Example  1.  The  population  of  a  certain 
city  in  1900  was  52,600,  and  in  1905  was  61,805.  Find  the 
gain  in  population  in  the  5  years.  What  was  the  gain  for  each 
100  of  the  population  during  the  5  years?  State  this  increase 
as  a  per  cent  of  the  population  in  1900.  What  was  the  average 
per  cent  of  increase  per  year? 

Discussion.  61,805  -  52,600  =  9205  =  gain  in  5  years. 
9205 -^526  =17. 5  =  gain  per  100  of  the  population.  Since, 
asking  for  the  per  cent  of  gain  is  the  same  as  asking  for  the 
number  of  increase  for  each  100  of  the  population,  therefore, 
stated  in  per  cent  this  is  17.5  %  of  the  population. 

1 7.5  %  -J-  5  » 3.5%  =  average  per  cent  of  increase  per  year, 
based  on  the  population  in  1900. 


PERCENTAGE  AND  APPLICATIONS  77 

Example  2.  In  a  certain  machine  f  of  the  energy  supplied 
to  the  machine  is  lost  in  friction  and  other  resistances.  What 
is  the  per  cent  of  efficiency?  If  f  of  the  loss  of  energy  is  in  a 
certain  part  of  the  machine,  what  per  cent  of  the  total  loss  is 
in  this  part? 

Discussion.  If  in  a  machine  it  is  known  that  -f  of  the 
energy  expended  is  wasted  in  frictional  and  other  resistances, 
we  say  that  40%  is  wasted,  meaning  that  T4<y°o  is  useless  for 
doing  work.  This  does  not  state  the  actual  numerical  amount 
of  energy  wasted;  all  it  tells  is  that  for  every  100  units  of  work 
expended  on  the  machine,  40  units  disappear.  Such  per- 
centages enable  comparisons  of  different  machines  to  be  made. 
If  one  machine  has  an  efficiency  of  60%  and  another  of  70%, 
we  know  that  the  second  is  10%  more  efficient  than  the 
first.  If  we  know  that  |  or  12|%  of  the  40%  loss  is  in  a 
certain  part,  this  gives  a  percentage  of  a  percentage.  The 
solution  of  the  problem  is: 

|  =  part  of  the  energy  lost, 

f  =  60%  =  efficiency  of  the  machine, 

I  of  $  = -sV  =  5  %  =  loss  in  the  particular  part  of  the  machine. 

62.  Averages  and  per  cent  of  error. — The  data  for  practical 
calculations  are* in  many  cases  either  the  result  of  measuring 
quantities,  or  of  experimental  observations,  and  in  each  case 
are  liable  to  error.  To  obtain  a  result  which  can  be  relied 
upon,  a  number  of  measurements  or  observations  are  taken  and 
the  average  or  mean  result  calculated. 

The  average,  or  mean  result,  is  obtained  by  adding  all  the 
measured  results  together  and  dividing  the  sum  by  the  number 
of  them.  This  average  is  accepted  as  the  best  approximation 
to  the  truth.  The  error  of  any  particular  observation  is 
obtained  by  finding  the  difference  between  it  and  the  average. 
This  error  can  often  be  most  conveniently  expressed  as  a  per 
cent,  and  is  spoken  of  as  the  per  cent  of  error.  We  always 
take  the  correct  value,  or  in  this  case  the  average  value,  as 
the  base. 

Example.  In  measuring  the  diameter  of  a  steel  rod  with 
a  micrometer,  the  separate  measurements  are:  0.3562  in., 
0.3569  in.,  0.3567  in.,  0.3570  in.,  and  0.3565  in.  Find  the 
average  measurement,  and  the  per  cent  of  error  in  the  largest 
and  the  smallest  measurements. 


78  PRACTICAL  MATHEMATICS 

Solution  ami  discussion. 

0.3562  in.  +  0.3569  in.  -I-  0.3567  in.  +  0.3570  in.  +  0.3565 
in.  =  1.7833  in. 

1.7833  in.  -*-  5  =  0.35666  in.  =  average. 

0.3570  in.  -  0.35666  in.  =  0.00034  in.  =  error  in  largest 
measurement. 

Using  the  formula  r  =  p  -r-  b  gives 

r  =  0.00034    in.  -*•  0.35666    in.  =  0.00095  +    =  0.095%  = 
per  cent  of  error  in  largest  measurement. 

0.35666  in. -0.3562  in.  =  0.00046  in.  =  error  in  smallest 
measurement. 

r  =  0.00046 in.  -5-  0.35666 in.  =  0.0013  =  0.13%  =  percent 
of  error  in  the  smallest  measurement. 

It  should  be  emphasized  that  the  per  cent  of  error  in  any 
measurement  is  always  found  by  using  the  correct  measure- 
ment as  the  base  and  the  error  as  the  percentage. 

63.  List  prices  and  discounts. — The  prices  of  machines 
and  materials,  printed  in  catalogs  and  price  lists,  are  usually 
subject  to  discounts.  Often  the  discount  is  so  large  that  the 
list  price  gives  no  idea  of  the  actual  cost.  In  preparing  an 
estimate,  it  is  necessary  to  know  what  discounts  are  given 
from  a  price  list. 

Discounts  are  usually  given  thus:  60%  and  10%  off  or 
simply  60  and  10  or  perhaps  "sixty  and  ten."  This  does  not 
mean  a  discount  of  70%,  but  that  a  discount  of  60%  is  first 
made  and  then  a  discount  of  10%  on  the  remainder.  Thus, 
if  the  list  price  is  $3.50  with  60%  and  10%  off,  we  find  60% 
of  $3.50,  which  is  $2.10.  Then  deduct  this  from  $3.50  leaving 
$1.40.  Now  get  10%  of  $1.40,  which  is  $0.14,  and  deduct  it 
from  $1.40,  leaving  $1.26  as  the  actual  cost. 

Similarly  we  may  have  discounts  of  40%,  10%,  and  4%, 
or  40,  10,  and  4  off.  These  are  deducted  in  turn  as  with  the 
two  discounts. 

EXERCISES  18 

1.  62%  of  2000  =  ?  Am.  1240. 

2.  What  is  |%  of  $28.80?  An*.  $0.108. 

3.  37i%  of  4000  =  ?  Ana.  1500. 

4.  300  is  1\%  of  what  number?  Ans.  4000. 
6.  What  per  cent  of  $104  is  $18.20?                                  Ans.  17$%. 
6.  What  per  cent  of  300  is  272?                                         Ana.  90|%. 


PERCENTAGE  AND  APPLICATIONS  79 

7.  The  indicated  horse-power  of  an  engine  is  10.6,  the  actual  effective 
horse-power  is  8.96.     What  per  cent  of  the  indicated  horse-power  is  the 
actual?  Ans.  84|$%. 

8.  A  bankrupt  has  $5760,  and  with  that  sum  can  pay  40%  of  his  debts; 
find  his  entire  indebtedness.  Ans.  $14,400. 

9.  For  collecting  a  bill  an  attorney  received  $2.52,  which  was  \\%  of 
the  bill;  find  the  amount  of  bill.  Ans.  $224. 

10.  A  milkman  sold  milk  at  7  cents  a  quart,  which  was  233|%  of  the 
cost;  find  cost  per  quart.  Ans.  3  cents. 

Suggestion.     233 \%  =\.     If  7  cents  is  I  of  the  cost,  what  is  the  cost? 

11.  What  is  the  net  price  per  barrel  of  oil,  the  list  price  of  which  is 
$18.00,  subject  to  a  discount  of  12^%  and  4%  off  for  cash? 

Ans.  $15.12. 

12.  A  tradesman  marks  his  goods  at  25%  above  cost  and  deducts  12% 
of  the  amount  of  a  customer's  bill  for  cash.     What  per  cent  does  he  make? 

Ans.  10%. 

Suggestion.  Suppose  the  cost  is  $20.  The  marked  price  is  25% 
above  $20  or  $25.  A  deduction  of  12%  on  $25  is  $3.  Hence  the  selling 
price  is  $25— $3  =$22.  The  gain  is  $22 -$20  =  $2.  What  per  cent  is 
$2  of  $20?  This  gives  the  gain  per  cent. 

13.  If  1225  pounds  of  coal  were  fired  to  a  boiler  and  152  pounds  were 
taken  out  of  the  ashpit  as  ash  and  waste,  what  per  cent  of  the  coal  was 
taken  from  the  ashpit.  Ans.   12.4  +  %. 

14.  The  weight  resting  on  the  drivers  of  a  locomotive  is  158,700  pounds. 
If  this  is  68.72%  of  the  total  weight;  find  the  weight  of  the  locomotive. 

15.  A  man  who  receives  42|  cents  an  hour  works  a  day  of  8  hours,  and 
4  hours  overtime  at  pay  for  time  and  a  half.     What  does  he  receive  in 
all?     What  per  cent  is  the  overtime  pay  of  the  total? 

Ans.  $5.95;426T%. 

15.  A  firm  increases  the  wages  of  its  employees  12|  %.  Find  the  wages 
of  a  man  who  was  getting  $3.40.  Of  a  boy  who  was  getting  $1.60  a  day. 
A  man  now  receives  $6.30  a  day.  What  did  he  receive  before  the 
increase?  Ans.  $3.82|;  $1.80;  $5.60. 

17.  A  tank  whose  capacity  is  168  gallons,  discharges  72  gallons  per 
hour,  which  equals  25%  less  than  it  receives.     In  what  time  will  it  be 
filled?  Ans.  7  hours. 

Suggestion.  If  72  gal.  is  25%  or  j  less  than  it  receives  per  hour,  72 
gal.  =  I  of  what  it  receives  per  hour.  Hence  it  receives  96  gal.  per  hour. 
Then  the  tank  receives  24  gal.  per  hour  more  than  it  discharges. 

18.  One  mill  is  gaged  at  767  barrels  of  flour  a  day,  which  equals  18% 
more  than  the  amount  for  another.     What  is  the  value  of  the  daily  out- 
put of  the  latter  at  $5  a  barrel?  Ans.  $3250. 

19.  The  usual  allowance  made  for  shrinkage  when  casting  iron  pipes  is 
£  in.  per  foot.     What  per  cent  is  this?  Ans.  1.04  +  %. 

20.  Find  the  cost  of  an  article  that  is  listed  at  80  cents,  40  and  6  off. 

Ans.  45.12  cents. 


80  PRACTICAL  MATHEMATICS 

21.  Find  the  cost  of  a  machine  quoted  at  $25.50,  40%  and  10%  off 
with  a  further  discount  of  4%  for  cash.  Ana.  $13.22. 

22.  When  rock  is  crushed  or  broken  into  fragments  of  nearly  uniform 
size  it  increases  in  hulk  and  has  voids,  or  inter-spaces,  of  from  30  to  55 
per  cent  of  the  whole  volume.     Find  the  number  of  cubic  yards  when 
crushed  occupied  by  1  cu.  yd.  of  solid  rock,  if  voids  are  (a)  30%;  (b) 
35%;  (c)  40%;  (d)  45%;  (e)  55%. 

Ans.  (a)  1.43;  (6)  1.54;  (c)  1.67;  (d)  1.82;  (e)  2.22. 
Suggestion.     The  volume  of  the  rock,  1  cu.  yd.,  is  30%  less  than  or 
70%  of  bulk  of  crushed  rock.     Using  formula 

b  =  p  •*•  r  gives  6  =  1  -f-  0.70  =  1.43  -  .  Ann. 

23.  A  team  of  horses  and  a  wagon  cost,  say  $400.     If  money  is  worth 
6%,  depreciation  in  value  of  team  and  wagon  is  25%  per  year,  teamster's 
wages  are  $2.00  per  day  while  working,  and  cost  per  month  for  keeping 
team  is  $18.50;  find  the  amount  that  should  be  charged  per  day  for  man 
and  team,  counting  250  days  actually  worked  per  year.       Ans.  $3.384. 

24.  In  the  preceding  problem  what  would  be  the  gain  per  year  from 
the  team  if  $5.00  a  day  was  charged  for  services,  deduction  being  made 
for  depreciation  in  value?  Ans.  $404. 

26.  In  estimating  the  amount  to  charge  per  day  for  the  use  of  a  steam- 
roller, a  contractor  has  the  following  data:  first  cost  of  steam-roller 
$3000;  money  worth  6%;  days  actually  worked  per  year,  100;  deprecia- 
tion in  value  of  the  machine,  $200  per  year.  Find  price  to  be  charged 
per  day  for  use  of  roller.  Ans.  $3.80. 

26.  The  actual  cost  of  removing  a  cubic  yard  of  rock  in  excavating  a 
certain  canal  is  $1.10.     What  price  should  be  put  in  the  estimate,  if 
12%  is  to  be  allowed  for  superintending,  and  10%  on  the  cost,  including 
superintending,  is  allowed  for  profit?  Ans.  $1.3552. 

27.  A  house  depreciates  in  value  each  year  at  the  rate  of  4%  of  its 
value  at  the  beginning  of  each  year,  and  its  value  at  the  end  of  two  years 
is  $6451.20.     Find  the  original  value.  Ans.  $7000. 

Suggestion.  $645 1. 20  -J-  96  =  $6720.  This  is  the  value  at  beginning  of 
second  year. 

28.  A  house  valued  at  $4000  rents  for  $27.50  per  month.     The  repairs 
on  house  each  year  amount  to  $40,  and  the  taxes  are  $17.50.     What 
interest  does  the  property  pay  on  the  investment,  no  allowance  being 
made  for  change  in  value  of  house?  Ans.  6H%- 

29.  In  making  a  certain  machine,  750  Ib.  of  iron  are  used  at  an  average 
cost  of  8  cents  per  pound.     There  are  used  in  the  work  on  the  machine, 
20  hr.  of  time  at  30  cents  per  hour,  7  hr.  at  60  cents  and  4  hr.  at  16  cents. 
If  20%  is  allowed  on  cost  as  profit,  what  is  the  selling  price  of  the  machine? 
What  will  it  be  listed  at  if  sold  at  30  and  5  off? 

Ans.  $85  nearly,  $127.82. 

30.  An  article  is  listed  at  $225,  and  sells  at  40  and  10  off.     How  will 
the  40%  discount  be  changed  to  offset  an  increase  of  15%  in  cost  of 
production?  Am.  31%,  or  better,  30%. 


PERCENTAGE  AND  APPLICATIONS  81 

Solution.  It  is  desired  to  find  a  first  discount  so  that  the  net  price, 
that  is,  the  price  after  the  discounts  are  made  from  the  list  price  will  be 
15%  more  than  the  net  price  when  discounts  of  40%  and  10%  are 
used. 

A  discount  of  40%  and  10%  off  from  $225  leaves  $121.50.  This  is 
the  old  net  price. 

$121.50  +  15%  of  $121.50  =  $139.725=new  net  price. 

$139.725-^0. 90  =  $155.25  =  price  after  new  first  discount  is  deducted 
from  list  price. 

$225  -$155.25  =  $69.75  =  amount  of  first  discount. 

$69.75 -^  $225  =0.31  =31%=  the  first  discount. 

31.  The  composition  of  white  metal  is  to  be  4  parts  by  weight  of  cop- 
per, 9  antimony,  and  97  tin.      Express  these  as  per  cents,  and  find  the 
weight  of  each  material  required  to  make  2376  Ib.  of  the  alloy. 

Ans.   Copper  3rV%,  86.4  Ib.;  antimony  81SI%,  194.4  Ib.; 
tin  88ft  %,  2095.2  Ib. 

32.  For  the  two  months  ending  Feb.  28,  1908,  there  were  exported 
from  the  United  States  $27,531,617  worth  of  iron  and  steel,  including 
machinery.     During  the  same  time  in  1909  it  was  $21,276,547.     Find 
the  decrease  per  cent.  Ans.  22.7  +  %. 

33.  The  output  of  Canadian  pig  iron  for  1908  was  563.672  tons,  a 
decrease  of  3%  from  1907.     What  was  the  output  in  1907? 

Ans.  581,105  tons. 

34.  Steel  billets  that  were  selling  at  $26  per  ton  dropped  to  $23  per 
ton.     What  is  the  per  cent  of  reduction?  Ans.   Ili73%. 

35.  The  water-power  in  use  in  the  United  States  is  5,300,000  horse- 
power.    The  undeveloped  is  8,100,000  horse-power.     What  per  cent  of 
the  total  water-power  is  developed  ?  Ans.  39.55  +  %. 

36.  A  ton  of  coal  from  the  Rock  Island  field  has  11.57%  moisture,  and 
6.27%  of  the  dry  coal  is  ash.     How  many  pounds  of  ash  in  a  ton  of  the 
coal?  Ans.   110.89+. 

37.  If  2.346  g.  of  an  ore  give  0.362  g.  of  copper,  what  per  cent  of  copper 
does  the  ore  contain?  Ans.   15.43  +  %. 

38.  2.3656  Kg.  of  ore  give  0.7  g.  of  gold  and  2.5  g.  of  silver.     Find  the 
per  cent  of  each.  Ans.  0.0296  -  % ;  0.1057  -  %. 

39.  A  merchant  buys  rubber  door  mats  at  $48.00  a  dozen  less  discounts 
of  40%,  15%,  and  5%.     What  should  he  sell  them  apiece  in  order  that 
he  may  make  35%?  Ans.  $2.62-. 

40.  lj-in.  basin  plugs  are  listed  by  the  jobber  at  $1.20  a  dozen.     The 
retailer  gets  discounts  of  50  and  10  off,  and  sells  them  at  15  cents  each. 
Find  his  gain  per  cent.  Ans.  233$%. 

41.  If  the  author  gets  10%  of  the  selling  price  of  a  book,  how  many 
hooks,  selling  at  75  cents  each,  must  be  sold  to  pay  the  author  $117.30? 

Ans.   1564. 

42.  In  a  compound  of  two  substances  A  and  B,  their  weights  are  in 
the  ratio  of  1.3498  to  1.     What  is  the  per  cent  of  each  in  the  compound? 

Ans.  57.44  +  %;  42.56-%. 


82  PRACTICAL  MATHEMATICS 

43.  Two  substances  A  and  B  form  a  compound  and  have  a  total 
weight  of  3.267  g.     If  the  compound  has  24.725%  of  A  and  75.275%  of 
B,  find  the  weight  of  each  substance  in  the  compound. 

Ana.  0.8078-  g. ;  2.4592  +g. 

44.  Find  the  cost  of  a  steam  boiler  listed  at  $500  subject  to  discounts 
of  40%,  10%,  and  7J%.  Ana.  $249.75. 

45.  Marshall  Field  and  Co.  quotes  an  article  of  silverware  at  $25  with 
discounts  of  40,  10,  5,  and  6%  off  in  10  days.     Find  net  cost  if  paid  in 
10  days.  Ana.  $12.06-. 

46.  The  recorded  measurement  of  a  city  block  is  528  ft.     By  chaining 
carefully  the  length  is  527.75  ft.     Find  the  per  cent  of  error  in  the  re- 
corded length.     How  wide  is  a  man's  lot  recorded  as  30  ft.? 

Ana.  0.047  +  %;  29. 986-  ft. 

47.  A  sample  of  nickel-steel  contained  24.51%  of  nickel  and  0.16%  of 
carbon.     How  much  of  each  nickel  and  carbon  in  2240  Ib.  of  nickel-steel? 

Ana.  549.024  Ib.;  3.584  Ib. 

48.  If  a  3i%  nickel-steel  rail  is  used  to  maintain  a  curve  in  a  street-car 
track,  it  lasts  three  times  as  long  as  carbon-steel.     How  much  will  be 
saved  per  ton  when  one  nickel-steel  rail  is  worn  out,  if  nickel-steel  costs 
$56  per  long  ton  and  carbon-steel  $28?     It  costs  $2.00  a  ton  for  laying, 
and  the  old  rails  are  worth  $16.00  per  ton,  besides  20  cents  a  pound  is 
realized  on  the  nickel.  Ana.  $15.68. 

49.  In  an  experiment  to  show  the  loss  of  pressure  for  different  kinds  of 
valves  in  water  pipes,  a  globe  valve  in  a  3-in.  pipe  caused  the  pressure  to 
fall  from  80  Ib.  to  41  Ib.  per  square  inch;  while  a  gate  valve  caused  a  loss 
of  pressure  of  4  Ib.  per  square  inch.     Find  (a)  the  per  cent  of  loss  for 
globe  valve,  (b)  for  gate  valve,  (c)  what  per  cent  loss  through  gate  valve 
is  of  loss  through  globe  valve.  Ana.  48J% ;  5% ;  10.26 -  %. 

60.  In  an  analysis  of  the  best  quality  of  crucible  cast  steel,  the  follow- 
ing was  found:  carbon  1.2%,  silicon  0.112%,  phosphorus  0.018%,  man- 
ganese 0.36%,  sulphur  0.02%,  iron  98.29%.     Find  the  number  of  pounds 
of  each  substance  if  the  total  weight  is  176.5  Ib. 

Ana.  2.118;  0.1977- ;  0.0318- ;  0.6354;  0.0353;  173.4S18+. 

61.  Find  the  cost  of  the  following  at  83%  discount: 

350  ft.,  8-in.  sewer  pipe  at  $0.50 

4  elbows  at  $2.00 

3  T  branches  at  $2.25 

4  traps  at  $6.60 

Ana.  $36.75. 

62.  The  mean  effective  pressure  on  the  piston  of  a  steam  engine, 
found  from  the  indicator  diagram,  was  59.75  Ib.  per  square  inch.     The 
boiler  pressure  was  87  Ib.  per  square  inch.     What  per  cent  of  the  boiler 
pressure  was  the  mean  effective  pressure?  Ana.  68.7  —  %. 

63.  The  grade  of  a  railroad  track  is  given  in  per  cent.     A  grade  of 
1%  is  a  rise  of  1  ft.  in  100  ft.     If  a  railroad  has  a  constant  grade  of  1J%, 
what  is  the  rise  in  3J  miles?  Ana.  231  ft. 


PERCENTAGE  AND  APPLICATIONS  83 

54.  The  total  rise  in  a  If  %  grade  is  43.6  ft.     Find  the  length  of  the 
track  having  this  grade.  Ans.  249  If  ft. 

55.  A  railroad  rises  112.7  ft.  in  3£  miles.     Find  the  average  grade. 

Ans.  0.61-%. 

56.  The  mechanical  efficiency  of  a  machine  is  the  relation  between  the 
work  put  into  the  machine  and  the  work  gotten  out  of  it.     Mechanical 
efficiency  is  usually  stated  as  a  per  cent.     Thus,  if  100  units  of  work  are 
put  into  a  machine  and  only  80  units  gotten  out  the  mechanical  efficiency, 
or  simply  the  efficiency,  is  80%.     What  is  the  efficiency  of  the  engine  of 
exercise  7? 

57.  At  what  advance  must  a  shopkeeper  mark  goods  costing  90  cents 
that  he  may  allow  a  20%  discount  and  yet  gain  25%?     Ans.  50f  cents. 

58.  A  man  purchases  ice  at  50  cents  per  100  Ib.     At  what  rate  must 
he  sell  it  after  it  has  lost  10%  of  its  weight  by  melting  to  gain  20%. 

64.  Interest.  —  Interest  is  money  that  is  paid  for  the  use  of 
money.  It  is  usually  reckoned  at  a  certain  rate  per  cent  per 
year.  The  base  on  which  the  interest  is  reckoned  is  called 
the  principal. 

In  percentage,  the  time  did  not  enter,  but  in  reckoning 
interest  the  time  has  to  be  taken  into  account.  The  interest 
on  a  sum  of  money  for  one  year  at  a  certain  rate  is  the  princi- 
pal multiplied  by  the  rate;  for  two  years  it  is  twice  as  much; 
and  for  any  period  of  time  it  is  the  interest  for  one  year  multi- 
plied by  the  time  in  years. 

If  p  stands  for  principal,  /  for  interest,  r  for  rate  per  cent, 
and  t  for  time  in  years,  the  interest  is  found  by  the  formula 


The  amount,  A,  is  the  principal  plus  the  interest. 

Many  short  methods  for  reckoning  interest  can  be  given, 
but  here  it  is  not  the  intention  to  enter  into  them. 

Example  1.  Find  the  interest  and  amount  of  $350  for 
5  years  at  6%. 

Z  =  pXrX£  =  $350X0.06X5  =  $105.00.  Ans. 
A  =  p+I  =  $350+$105.00  =  $455.00.  Ans. 

Example  2.  Find  the  interest  on  $750  for  2  yr.  7  mo. 
at  8%. 

Here  the  time  is  f  £  years,  since  in  getting  the  time  in  years 
we  use  12  months  for  a  year,  30  days  for  a  month,  and  360 
days  for  a  year. 

/  .  /  =  $750  X  0.08  X  H  =  $1  55.00.  A  ns. 


84  PRACTICAL  MATHEMATICS 

It  is  usually  best  to  use  cancellation. 
750X8X31 


Example  3.     Find  the  interest  on  $375  for  2  yr.  5  mo.  15  da 
at  5%. 

Here  the  time  is  f|  ft  years. 

/.  7  =  $375X0.05X|f!  =  $46.09.  Ans. 

EXERCISES  19 

Find  the  interest  and  amount  of  each  of  the  following: 

1.  $700  for  3  yr.  at  8%.  Ans.  $168.00;  $868.00. 

2.  $14.30  for  2  yr.  9  mo.  at  8%.  Ans.  $3.15;  $17.45. 

3.  $245.60  for  2  yr.  7  mo.  21  da.  at  8%.         Ans.  $51.90;  $297.50. 

4.  $436.75  for  1  yr.  2  mo.  15  da.  at  5%.         Ans.  $26.39;  $463.14. 
6.  $325.25  for  2  yr.  9  mo.  12  da.  at  6J%.        Ans.  $58.84;  $384.09. 

6.  $87.50  lor  3  yr.  3  mo.  at  7%.  Ans.  $19.91;  $107.41. 

7.  $480  for  6  yr.  3  mo.  at  15%.  Ans.  $450;  $930. 

8.  $18.20  for  9  yr.  9  mo.  9  da.  at  5J%.  Ans.  $10.23;  $28.43. 

9.  A  note  for  $225  at  6%  runs  for  9  mo.  What  is  the  amount  of  the- 
note  when  due?  Ans.  $235.13. 

10.  A  note  for  $390.00  at  7%  runs  for  3  yr.  6  mo.     What  is  the  amount 
due?  Ana.  $485.55. 


CHAPTER  VII 
RATIO  AND  PROPORTION 

65.  Ratio. — There  are  several  ways  of  stating  the  relation 
of  one  quantity  to  another.  If  the  size  or  magnitude  of 
the  quantities  are  thought  of,  a  very  convenient  way  of  com- 
paring them  is  to  state  the  ratio  of  one  to  the  other. 

The  ratio  of  one  number  to  another  is  the  quotient  of  the 
first  number  divided  by  the  second. 

fB»/» 

Thus,  the  ratio  of  $6  to  $2  is  3,  and  may  be  stated  in  the  form  -^  or 

V™ 

$6  :  $2.     In  either  case  it  is  read  "the  ratio  of  $6  to  $2." 

From  the  idea  of  a  ratio  it  is  evident  that  we  can  state 
a  ratio  between  two  magnitudes  only  when  the  magnitudes 
are  alike.  That  is,  a  ratio  cannot  be  stated  between  such 
quantities  as  dollars  and  bushels. 

The  two  numbers  used  in  a  ratio  are  called  the  terms  of 
the  ratio.  The  first  one  is  named  the  antecedent  and  is  the 
dividend;  the  second  is  named  the  consequent  and  is  the 
divisor. 

The  ratio  2  : 3  is  the  inverse  of  the  ratio  3  : 2. 

Since  a  ratio  in  the  form  4:3  is  an  indicated  division 
or  a  fraction,  the  principles  applying  in  division  or  to  a  fraction 
likewise  apply  to  a  ratio. 

The  expressions  "in  the  same  ratio  as,"  "in  the  same  pro- 
portion," "proportionally,"  and  "pro  rata"  all  have  practically 
the  same  meaning. 

When  it  is  said  that  $20  is  divided  between  two  men  in  the  ratio  of 

2  to  3,  it  is  meant  that  one  gets  $2  as  often  as  the  other  gets  $3.     That 
is,  of  each  $5,  one  gets  $2  and  the  other  $3.     Hence  one  gets  |  of  $20  or 
$8,  and  the  other  gets  |  of  $20  or  $12. 

EXERCISES  20 

1.  Find  the  value  of  the  following  ratios:  8:2;  9:4;  17:2J;  44  hours: 

3  hours;  7bu.:2  bu.;  4^:3£;  9*:  16. 

2.  A  room  is  16  ft.  by  12  ft.     What  is  the  ratio  of  its  length  to  its 
width? 

85 


80  PRACTICAL  MATHEMATICS 

3.  Two  gear  wheels  have  80  teeth  and  30  teeth  respectively.     What  is 
the  ratio  of  the  numbers  of  teeth? 

4.  One  city  has  a  population  of  8000  and  a  second  a  population  of 
20,000.     What  is  the  ratio  of  their  populations?     What  part  is  the  first 
of  the  second?     What  per  cent?     How  many  times  as  large  as  the  first 
is  the  second?     What  difference  is  there  in  the  ideas  involved  in  the 
questions? 

6.  W'rite  the  inverse  ratios  to  the  following:  7:2;  9:2J;  10  ft.  :90  ft,; 
23J:2t. 

6.  Divide  $50  between  A  and  B  in  the  ratio  of  3:7. 

7.  A  man  rode  250  miles  partly  by  rail  and  partly  by  boat.     What 
distance  did  he  travel  by  each  if  their  ratio  is  as  3  to  2? 

,4ns.   150  mi. ;  100  mi. 

8.  Fifty-one  students  entered  a  class  and  33  of  them  finished  the  work. 
What  per  cent  finished?     What  is  the  ratio  of  the  number  that  finished 
to  the  whole  number? 

9.  A  worm  wheel  makes  6  turns  per  minute  and  the  worm  180  turns 
per  minute.     What  is  the  ratio  of  the  reduction  of  speed ?     Ans.  30tol. 

66.  Proportion. — A  proportion  is  a  statement  of  equality 
between  two  ratios. 

Thus,  2:3=4:6  and  4  men: 8  men  =$6: $12,  are  proportions. 

The  first  and  last  terms  of  a  proportion  are  called  the 
extremes.     The  second  and  third  terms  are  called  the  means. 

In  the  first  proportion  above,  2  and  6  are  the  extremes  and  3  and  4 
the  means. 

By  inspecting  several  proportions  the  following  principles 
will  be  evident: 

(1)  The  product  of  the  means  of  any  proportion  is  equal  to  the 
product  of  the  extremes. 

(2)  The  product  of  the  two  means  divided  by  either  extreme 
gives  the  other  extreme. 

(3)  The  product  of  the  two  extremes  divided  by  either  mean 
gives  the  other  mean. 

Example   1.     Find   the   value   of   h   from   the   proportion 
25  :  100  =  7  :h. 

100X7 

Solution.     Applying  principle  (2),  h  =  -  -~= —  =  28.  Ans. 

£\t 

Example  2.     If  15  tons  of  coal  cost  $63  what  will  27  tons 
cost  at  the  same  rate  per  ton  ? 

Solution.     Since  the  same  relation  holds  between  the  cost 


RATIO  AND  PROPORTION 


87 


prices  as  between  the  amounts  of  coal,  the  ratio  of  15  tons  to 
27  tons  must  equal  the  ratio  $63  to  the  cost  of  27  tons. 
Let  x  stand  for  the  number  of  dollars  27  tons  cost,  and  we 
can  state  the  proportion, 

15  :  27  =  63  :  x. 
27X63 


15 


=  113.40. 


.'.  27  tons  cost  $113.40.  Ans. 

Example  3.  If  25  men  can  do  a  piece  of  work  in  30  days, 
in  how  many  days  can  35  men  do  the  same  work? 

Solution.  It  is  evident  that  35  men  can  do  the  work  in  less 
time  than  25  men,  hence  the  ratio  of  the  number  of  days  is 
equal  to  the  inverse  ratio  of  the  number  of  men.  Using  x 
for  the  number  of  days  required, 

35  :25  =  30 
25X30      • 


x. 


X  — 


35 


.'.  35  men  can  do  the  work  in  21-f  days.  Ans. 
Example  4.     An  inclined  plane  as  shown  in  the  figure  rises 
38  ft.  in  100  ft.,  find  the  height  h  it  will  rise  in  28  ft. 


Solution.     Here  the  proportion  is 

100  :  28  =  38  :  h. 

,     28X38 
..h=~m-  =  10.64. 

.'.  the  rise  in  28  ft.  is  10.64  ft.  Ans. 

The  proportion  could  as  well  be  stated  100  : 38  =  28  :  h. 

Definition.     If  the  rise  of  a  road  bed  is  h  ft.  in  100  ft.,  the 

grade  of  the  road  is  r^'  or  the  ratio  of  the  rise  to  the  horizontal 


88  PRACTICAL  MATHEMATICS 

distance.     Thus,  if  a  road  rises  3  ft.  in  100  ft.  the  grade  is 

JL    _  QC7 
100  " 

Example  5.     What  is  the  grade  of  a  road  bed  that  rises 
1.2  ft.  in  a  horizontal  distance  of  40  ft.? 
Solution.    Let  h  stand  for  the  number  of  feet  rise  in  100  ft. 

It  is  evident  that  the  ratio  T  ~  =  the  ratio  —•  But  - 

100  40  40 

0.03.     .*.  the  grade  is  0.03  or  3%,  Ans. 

Example  6.  If  a  bell  metal  is  25  parts  copper  to  12  parts 
tin,  what  is  the  weight  of  each  in  a  bell  weighing  1850  lb.? 

Solution.  The  ratio  of  the  number  of  parts  of  each  metal 
to  the  whole  number  of  parts  equals  the  ratio  of  the  weight  of 
each  metal  to  the  whole  weight.  Use  c  to  stand  for  the  number 
of  pounds  of  copper,  and  t  for  the  tin.  Then  we  have 

25  :  37  =  c  :  1850, 
and  12  : 37  =  t  :  1850. 

25X1850     19_n 
.  .  c  = — =  1250, 

12X1850 

and  t  = 5= =  600. 

o7 

.'.  weight  of  copper  is  1250  lb.  and  tin  is  600  lb.  Ans. 

EXERCISES  21 

Find  the  value  of  the  letter  in  the  exercises  1  to  6. 

1.  17:45  =  14:*.  Ans.  z=37iV. 

2.  3|:9j=6:x.  Ans.  z  =  15H- 

3.  16|:29J=50&:z.  Ans.  z  =  88*. 

4.  3:z  =  5:25.  Ans.  x-15. 
6.  75:85  =x:  170.  Ans.  a:  =  150. 

6.  r:  11  =  17: 121.  Ans.  r  =  l^. 

7.  If  a  train  travels  378  miles  in  11  hours,  how  far  will  it  travel  in  17 
hours?  Ans.  684ft  miles. 

8.  If  10  men  can  do  a  piece  of  work  in  20  days,  how  long  will  it  take 
25  men  to  do  it?  Ans.  8  days. 

9.  If  a  ship  sails  256  miles  in   11'  hours,  how  far  will  it  sail  a  the 
same  rate  in  179  hours?  Ans.  3984lj  miles. 

10.  The  roof  of  a  house  rises  2  ft.  in  a  run  of  3  ft.,  how  far  will  it  rise 
in  a  run  of  20  ft.  ?  Ans.  13  ft.  4  in. 

11.  A  road  bed  rises  2\  ft.  in  200  ft.,  what  is  the  grade?    In  how  many 
feet  will  it  rise  1  ft.  ?  Ans.  1 J  % ;  80. 


RATIO  AND  PROPORTION  89 

12.  If  16f  tons  of  hay  cost  $61.875,  find  the  cost  of  28  tons  at  the 
same  rate.  Ans.  $105. 

13.  The  mixture  for  a  casting  has  4  parts  of  copper,  3  parts  lead,  and 
2  parts  tin.     How  many  pounds  of  each  in  a  casting  weighing  96  Ib.  ? 

Ans.  42f  ;  32;  21J. 
Work  the  following  4  exercises  by  proportion. 

14.  What  per  cent  is  59.1  of  51.3? 

Solution.  100%  stands  for  the  base,  then,  using  x  for  the  number  of 
per  cent  required,  the  proportion  is 

51.3  :  51.9  =  100  :z. 
.  59.1X100 

51.3 
.'.59.1  is  115.2%  of  51.3. 

15.  46  is  what  per  cent  of  79?  Ans.  58.23  -  %. 

16.  146  is  17%  of  what  number?  Ans.  858.82+. 

17.  3%  of  a  number  is  426,  what  is  the  number?  Ans.   14,200. 

67.  Measuring  heights.  —  There  are  several  methods  for 
determining  the  height  of  a  standing  tree.  One  of  the  simplest 
is  to  measure  the  shadow  of  the  tree  and  the  shadow  of  a 
straight  pole  of  known  length  set  upright  in  the  ground. 
Then  if  H  stands  for  the  height  of  the  tree,  h  for  the  height 
of  the  pole,  S  for  the  length  of  the  shadow  of  the  tree,  s 
for  that  of  the  pole,  we  have  the  proportion 

s:S=h:H. 

EXERCISES  22 

1.  Find  the  height  of  a  tree  that  casts  a  shadow  115  ft.  long  when  a 
pole  8  ft.  high  casts  a  shadow  of  5  ft. 
Solution.    5:8  =  115:  H. 


=  = 

5 

/.  height  of  tree  is  184  ft.  Ans. 

2.  Find  the  height  of  a  church  steeple  that  casts  a  shadow  84  ft.  long 
when  a  pole  1  1  ft.  long  casts  a  shadow  of  7  ft.  9  in.  Ans.  1  19  ft.  nearly. 

The  two  following  methods  with  the  figures  are  given  in  Bulletin  36 
of  the  Bureau  of  Forestry,  U.  S.  Department  of  Agriculture. 

(1)  A  method  used  when  the  sun  is  not  shining  is  to  set  two  poles  in 
a  line  with  the  tree  as  shown  in  Fig.  16.  From  a  point  S  on  one  pole 
sight  across  the  second  pole  to  the  base  and  to  the  top  of  the  tree.  Let 
an  assistant  note  the  points  a  and  b  where  the  lines  of  vision  cross  the 
second  pole  and  measure  the  distance  between  these  points,  ab,  also 
measure  the  distance  from  the  sighting  point  on  the  first  pole  to  the  base 


90  PRACTICAL  MATHEMATICS 

of  the  tree,  SB,  and  to  the  lowest  point  on  the  second  pole,  56.     Then  the 
following  proportion  is  true: 

56 :  SB  =  06 :  AB. 


3.  Find  the  height  of  a  tree  when  56  =  6  ft.,  5fi=40  ft.,  and  06  =  9  ft. 

Ans.  60  ft. 

(2)  Another  method  sometimes  used  is  as  follows:  The  observer  walks 
on  level  ground  to  a  point  A  at  a  convenient  distance  AD  from  the  foot 
of  the  tree.  He  then  lies  on  his  back  as  shown  in  Fig.  17.  An  assistant 


RATIO  AND  PROPORTION 


91 


notes  on  an  upright  staff  erected  at  his  feet  the  exact  point  C  where 
his  line  of  vision  to  the  top  of  the  tree  E  crosses  the  staff.  The  height 
of  the  staff  BC  is  measured,  and  his  own  height  AB  from  his  feet  to  his 
eyes,  then  the  following  proportion  is  true: 

AB:BC  =  AD:DE. 

4.  Find  the  height  of  the  tree  DE  if  AB=5±  ft.,  BC  =  8  ft.,  and 
AD  =  90  ft.  Ans.  131  ft,  nearly. 

68.  The  lever. — A  stiff  bar  or  rod  supported  at  some 
pivotal  point,  about  which  it  can  move  freely,  is  called  a  lever. 
The  pivotal  point  is  called  the  fulcrum.  The  lever  enters 
in  one  form  or  another  into  many  mechanical  devices. 


-D- 


(a) 


In  Fig.  18,  F  stands  for  fulcrum,  W  for  the  weight  lifted, 
P  for  the  force  that  does  the  lifting,  D  for  the  distance  from 
the  fulcrum  to  the  point  of  application  of  the  force,  and  d 
for  the  distance  from  the  fulcrum  to  the  point  where  the  weight 
is  attached.  In  all  possible  relations  of  the  fulcrum,  weight, 
and  force  the  following  proportion  holds: 

P:W  =  d:D. 

That  is,  the  applied  force  is  to  the  weight  inversely  as  their 
distances  from  the  fulcrum.  This  means  that  a  small  force 
will  balance  a  larger  weight  only  if  the  weight  is  nearer  the 
fulcrum  than  the  force. 


'.'•J 


PR  A CT1CAL  MA  Til K\tA  TICS 


EXERCISES  23 

1.  Given  P  - 150  Ib.,  D-12*  ft.,  d-lj  ft,  find  W.      Ans.  1250  Ib. 

2.  Given  P -200  Ib.,  D-9J  ft,  TF-775  Ib.,  find  d. 

Ans.  2ft.  6.2-  in. 

3.  Given  P  =  160  Ib.,    TF=900 
Ib.,  rf-ljft.,  find  D. 

Ana.  7  ft  f  in. 

4.  Given  IF  =  160  Ib.,  D  =  3$  ft, 
rf  =  8J  ft,  find  P. 

Ans.  381  Ib.  nearly. 
6.  In  a  wire  cutter  the  wire  is 
placed  i  in.  from  the  fulcrum  and 
the  pressure  of  the  hand  is  7  in. 
from  the  fulcrum.  Find  the  re- 
sistance of  the  wire  if  the  hand 
exerts  a  force  of  40  Ib. 

Ans.  560  Ib. 

6.  In  pulling  a  nail  from  a  board 
with  a  hammer  as  shown  in  Fig. 
19,  find  the  resistance  of  the  nail 

at  the  start  if  P  =  50  Ib.,  D  =  10  in.,  and  d  =  H  in.  Ans.  333  J  Ib. 

7.  In  the  ordinary  steel-yard,  Fig.  20,  what  must  be  the  weight  P  to 


FIG.  20. 

balance  a  weight  W  of  17  J  Ib.  if  it  is  1$  in.  from  fulcrum  to  application 
of  W  and  8J  in.  from  application  of  P  to  fulcrum?    Ana.  2  Ib.  9.2—  oz. 

8.  If  the  steel-yard  is  turned  over  so  that  the  distance  from  the  ful- 
crum to  W  is  i  in.,  what  weight  W  will  1J  Ib.  at  P  balance  when  P  is 
20 J  in  from  the  fulcrum?  Ans.  62 i  Ib. 


RATIO  AND  PROPORTION 


93 


69.  Hydraulic  machines. — A  principle  known  as  Pascal's 
Law  states  that  pressure  exerted  on  a  liquid  in  a  closed  vessel 
is  transmitted  equally  and  undiminished  in  all  directions. 

In  Fig.  21,  if  the  area  of  a  is  1 
sq.  in.  then  a  pressure  of  1  Ib.  at 
a  gives  a  pressure  of  1  Ib.  on  each 
square  inch  of  the  surface  of  C.  If 
the  area  of  the  top  of  C  is  100  sq. 
in.  then  a  pressure  of  1  Ib.  at  a 
will  lift  a  weight  of  100  Ib.  at  A. 

If  a,  A,  p,  and  P  are  the  areas 
and  pressures  respectively  then  we 
have  the  proportion 

a  :  A=p  :P. 

EXERCISES  24 

1.  A  pressure  of  5  Ib.  on  the  cork  of  a 
jug  filled   with  water  gives   how   many 
pounds  pressure  tending  to  force  out  the 
bottom    of  the  jug?     The   area   of   the 

cork  is  1|  sq.  in.  and  the  area  of  the  bottom  is  245.6  sq.  in. 

Ans.  818|  Ib. 

2.  A  hydraulic  lifter  used  to  raise  heavy  weights  has  the  pressure  ap- 
plied to  a  piston  having  an  area  of  5  sq.  in.  by  a  lever.     From  the  ful- 
crum to  the  point  attached  to  the  small  piston  is  4  in.,  and  to  the  point 
where  a  force  of  100  Ib.  is  applied  is  22  in.     Find  the  weight  that  can  be 
raised  on  a  piston  having  an  area  of  75.6  sq.  in. 

Solution.     Let  x  =  pressure  in  pounds  applied  on  small  piston. 
Then  1 00 :  x  =  4 : 22.     From  which  x  =  550. 
And  0.5: 75.6  =  550:  P. 


FIG.  21. 


550  X  75.6 
0.5 


=  82,160. 


Hence  a  weight  of  82,160  Ib.  can  be  raised. 

3.  A  supply  pipe  for  a  14-in.  plunger  hydraulic  elevator  piston  is  1J 
sq.  in.  in  area,  and  the  pressure  in  the  supply  pipe  is  pumped  up  to  150 
Ib.  per  square  inch.  What  is  the  total  pressure  on  the  14-in.  plunger  if 
it  has  an  area  of  153.94  sq.  in.?  Ans.  23,091  Ib. 


CHAPTER  VIII 
DENSITY  AND  SPECIFIC  GRAVITY 

70.  Density. — Experience   tells   us   that   some   bodies   are 
heavier  than  others;  that  is,  of  two  bodies  of  the  same  size, 
one  weighs  more  than  the  other.     Take  a  cubic  foot  of  metal 
and  one  of  wood;  suppose  the  metal  weighs  500  Ib.  and  the 
wood  50  Ib.,  then  the  metal  is  ten  times  as  heavy  as  the  wood, 
or  the  ratio  of  their  densities  is  as  10  to  1 .     We  also  say  that  the 
density  of  the  metal  is  500  Ib.  per  cubic  foot. 

Water  has  a  density  of  about  62.5  Ib.  per  cubic  foot.  In 
the  metric  system  the  density  of  water  under  standard  con- 
ditions is  one  gram  per  cubic  centimeter. 

The  density  of  a  body  is  its  mass  per  unit  volume.  For 
our  purpose  the  mass  is  the  same  as  the  weight.  Strictly 
speaking,  the  weight  of  a  body  near  the  earth  is  the  force 
with  which  the  earth  attracts  the  mass  of  the  body. 

71.  Specific  gravity. — The  term  specific  gravity  is  used  for 
the  ratio  of  the  densities  of  two  bodies.     Thus,  the  specific 
gravity  of  the  metal  with  reference  to  the  wood  is  10,  which 
means  that  the  metal  is  ten  times  as  heavy  as  the  wood. 
It  should  be  carefully  noticed  that  the  specific  gravity  of  a 
substance  is  an  abstract  number,  that  is,  a  number  with  no 
name  attached. 

72.  Standards. — For  convenience  the  standard   to  which 
other  substances   are  referred,   in   stating  specific  gravities, 
is  water  for  solids  and  liquids. 

RULE.  The  specific  gravity  of  a  substance  is  obtained  by 
finding  the  weight  of  a  certain  volume  of  it  and  dividing  thin 
weight  by  the  weight  of  the  same  volume  of  the  standard. 

Thus,  to  find  the  specific  gravity  of  a  stone  it  is  necessary 
to  find  its  weight,  and  the  weight  of  an  equal  volume  of  water. 
The  weight  of  the  stone  divided  by  the  weight  of  the  water 
gives  the  specific  gravity  of  the  stone. 

The  specific  gravity  of  any  other  body  could  be  found  in 
the  same  manner.  Some  difficulty  might  be  found  in  doing 

94 


DENSITY  AND  SPECIFIC  GRAVITY  95 

the  weighing,  but  a  little  ingenuity  will  devise  a  plan.     Vari- 
ous methods  for  doing  the  weighing  are  discussed  in  physics. 

Water  is  taken  as  the  standard  because  of  its  abundance. 
All  substances  can  be  referred  to  it,  but  gases  are  usually 
compared  with  air  or  hydrogen  gas. 

If  w  stands  for  the  weight  of  the  body  whose  specific  gravity 
is  to  be  found,  s  the  weight  of  the  same  volume  of  the  stand- 
ard, and  g  for  the  specific  gravity,  the  rule  may  be  stated  as  a 
formula  : 

w  -T-  s  =  g. 

73.  Use.  —  Tables  of  the  specific  gravities  of  the  various 
substances  are  given  for  use  in  making  computations.  In 
Table  VIII  are  given  the  specific  gravities  of  a  few  of  the  more 
common  substances. 

If  it  is  required  to  find  the  weight  of  a  block  of  iron  2  ft. 
by  3  ft.  by  1  ft.  we  could  find  the  number  of  cubic  feet  in  the 
block  which  is  6.  This  times  the  weight  of  a  cubic  foot  of 
water  gives  the  weight  of  an  equal  volume  of  water,  or  62.5X6 
=  375  Ib.  The  weight  of  the  water  multiplied  by  the  specific 
gravity  of  iron  gives  the  weight  of  the  iron,  or  375  Ib.  X7.2 
=  2700  Ib. 

In    terms    of    the    letters    already    used,    since    w  +  s  =  g, 


Example  1.     Find  the  specific  gravity  of  a  rock  if  1  cu.  ft. 
of  it  weighs  182  Ib. 

Solution.     Since  water  weighs  62.5  Ib.  per  cubic  foot  the 
specific  gravity  of  the  rock  is  found  thus: 
182  lb.^62.5  lb.  =  2.912. 
.'.  specific  gravity  of  the  rock  is  2.912. 

Example  2.     How  many  cubic  inches  are  there  in  1  Ib.  of 
cork,  if  its  specific  gravity  is  0.24? 

Solution.     Since  1728  cu  in,  of  water  weigh  62.5  Ib. 

1728  cu.  in.  of  cork  weigh  0.24X62.5  Ib. 


.'.  1  cu.  in.  of  cork  weighs  —— 7-700-"  -  Ib. 


A  nrl   1  -• _"••"•*  ^"*-" 1728        _  1  i  K  o 

1728       ~  0.24X6275" 

.'.there  are  115.2  cu.  in.  in  1  Ib.  of  cork. 


96  PRACTICAL  MATHEMATICS 


EXERCISES  26 

1.  Find  the  weight  of  176  cu.  in.  of  copper.  An*.  56+  Ib. 

2.  Find  the  weight  of  37  cu.  ft.  of  cast  iron.  Ans.  16,650  Ib. 

3.  A  stone  weighs  3  Ib.  in  air  and  1.75  Ib.  in  water.     Find  its  specific 
gravity.  Ans.  2.4. 

4.  What  is  the  specific  gravity  of  a  substance  40  cu.  in.  of  which 
weighs  6  Ib.?  Ans.  4.1 47  +  . 

6.  Two  cubic  feet  of  cast  iron  immersed  in  water  weigh  how  much? 
Solution.     From    Table   VIII,   2   cu.   ft.   of   cast  iron   weigh   in   air 
2X450  Ib.  =9001b. 

2  cu.  ft.  of  water  weigh  2X62.5  Ib.  =  125  Ib. 
Weight  of  iron  in  water  =  900  Ib.  -125  Ib.  =775  Ib. 

6.  A  piece  of  metal  weighing  243  Ib.  floats  in  mercury  (s.  g.  13.6)  with 
1*7  of  its  volume  immersed.     Determine  the  volume  and  the  specific 
gravity  of  the  metal.  Ans.  s.  g.  =7.2;  Vol.  =933.1  in.* 

7.  The  specific  gravity  of  ice  is  0.92,  of  sea  water  1.025.     What  part 
of  an  iceberg  is  below  the  surface  of  the  water  when  floating? 

Ans.  0.8975. 

8.  A  balloon  containing  10,200  cu.  ft.  will  lift  how  great  a  weight  if 
filled  with  hydrogen  gas?  Ans.  Less  than  756.5  Ib. 

9.  An  irregular  shaped  mass  of  iron  (s.  g.  7.22)  weighed  in  air  126  Ib. 
Find  its  volume.     What  would  be  its  weight  if  immersed  in  water? 

Ans.  482.5-  in.»;  108.55-  Ib. 

10.  A  pond  f  acre  in  area  is  frozen  over.     Find  the  weight  in  tons  of 
the  ice  if  it  is  3J  in.  thick  and  the  specific  gravity  of  ice  is  0.92. 

Ans.  273.95  tons. 

11.  Find  the  weight  of  a  cubic  meter  of  iron  (s.  g.  7.22)  in  kilograms. 
What  is  the  weight  in  pounds?  Ans.  7220  Kg. ;  15935.6  Ib. 

12.  Find  the  number  of  liters  in  a  vat  2  m.  X75  cm.  X50  cm.     Also 
find  the  weight  in  Kg.  of  the  sulphuric  acid  (s.g.  1.84)  required  to  fill  it. 

Ans.  7501.;  1380  Kg. 

13.  Find  the  value  of  17  1.  of  sulphuric  acid  at  5  cents  per  Kg. 

Ans.  $1.56. 

14.  Mercury  weighs  13.596  times  as  much  as  water  at  its  greatest 
density.     What  is  the  pressure  per  square  centimeter  of  a  column  of 
mercury  76  cm.  high?  Ans.  1033.296  g. 

16.  A  column  of  mercury  how  high  would  cause  a  pressure  per  square 
inch  equal  to  14.7  Ib.?  Ans.  29.89+  in. 

16.  A  tank  1.85  m.  long,  1.35  m.  wide,  and  85  cm.  deep  is  filled  with 
sea  water  (s.  g.  1.025).     What  is  the  weight  of  the  water? 

Ans.  2175.95-  Kg. 

17.  Sandstone  of  specific  gravity  2.5  is  crushed.     Find  the  weight  of 
1  cu.  yd.  of  the  crushed  stone  if  the  voids  are  35%.     (See  Ex.  22,  p.  80). 

Solution.     If  35%  are  voids,  65%  is  rock. 
Weight  =  0.65  X27  X62.5  X2.5  =-2742  +  Ib. 


DENSITY  AND  SPECIFIC  GRAVITY  97 

18.  Granite  of  specific  gravity  2.8  is  crushed.     Find  the  weight  of 
1  cu.  yd.   of  the  crushed  rock  if  voids  are  40%.         Ans.  2835  Ib. 

19.  A  casting  of  iron  when  immersed  in  water  displaces  2  quarts;  find 
the  weight  of  the  casting.  Ans.  30  Ib. 

20.  An  irregular  shaped  steel  forging  was  found  to  displace  6.75  quarts 
of  water;  find  the  weight  of  the  forging.     (Use  s.  g.  of  steel  =  7.85). 

Ans.   Ill  Ib.  nearly. 

21.  A  wooden  pattern  for  a  casting  weighs  2f  pounds.     An  aluminum 
casting  is  to  be  made.     Find  the  weight  of  the  casting  if  the  specific 
gravity  of  the  wood  is  0.52  and  that  of  the  aluminum  is  2.6. 

Ans.   13f  Ib. 


CHAPTER  IX 
POWERS  AND  ROOTS 

74.  Powers. — When  we  have  several  numbers  multiplied 
together,  as  3X4X6  =  72,  we  call  the  numbers  3,  4,  and  6, 
factors  and  72  the  product.     If  now  we  make  all  the  factors 
alike,  as  3X3X3X3=81,  we  call  the  product  by  the  special 
name  power.     We  say  81  is  a  power  of  3,  and  3  is  the  base  of 
the  power. 

A  power  is  a  product  obtained  by  using  a  base  a  certain 
number  of  times  as  a  factor. 

If  the  base  is  used  twice  as  a  factor  the  power  is  called  the 
second  power;  three  times  as  a  factor,  the  third  power;  and 
so  on  for  any  number  of  times. 

75.  Exponent   of   a   power. — Instead   of   3X3X3X3,    we 
may  write  34.     The  small  figure,  placed  at  the  right  and  above 
the  base,  shows  how  many  times  the  base  is  to  be  used  as  a 
factor,  and  is  called  an  exponent. 

The  exponent  of  a  power  is  a  number  placed  to  the  right 
and  above  a  base  to  show  how  many  times  the  base  is  used  as 
a  factor. 

It  should  be  noted  that  the  use  of  the  exponent  gives  us  a 
short  concise  way  of  writing  a  continued  product  where  the 
factors  are  all  alike. 

76.  Squares,  cubes,  involution. — The  second  power  of  a 
number  is  called  the  square  of  the  number,  as  32. 

The  third  power  of  a  number  is  called  the  cube  of  the  number, 
as  5s. 

The  higher  powers  have  no  special  names.  34  is  called  the 
fourth  power  of  3,  57  the  seventh  power  of  5,  etc. 

Involution  is  the  process  of  finding  the  powers  of  numlx?rs. 

98 


POWERS  AND  ROOTS  99 

EXERCISES  26 

1.  Find  the  square  of  7,  of  27,  of  92,  of  736.     Find  the  square  of  the 
square  of  3,  of  7,  of  10. 

2.  Find  the  cube  of  7,  of  8.     Find  the  square  of  the  cube  of  3. 

Arcs.  343;  512;  729. 

3.  Find  the  fourth  power  of  5.     What  is  the  difference  between  the 
fourth  power  of  a  number  and  the  square  of  the  square  of  the  same 
number? 

4.  Find  values  of  the  following:  (a)  7922,  (6)  353,  (c)  34,  (d)  216. 

Ans.   (a)  627,264,  (6)  42,875,  (c)  81,  (d)  65,536. 

77.  Roots. — If  we  take  9  and  separate  it  into  the  two  equal 
factors  3  and  3,  that  is,  9  =  3X3,  then  one  of  these  factors, 
3,  is  called  the  square  root  of  9.     The  process  is  just  the  inverse 
of  that  by  which  the  power  is  found.     Similarly  64=4X4X4, 
and  we  say  4  is  the  cube  root  of  64. 

The  square  root  of  a  number  is  one  of  the  two  equal  factors 
into  which  a  number  is  divided. 

The  cube  root  is  one  of  the  three  equal  factors  into  which 
a  number  is  divided;  the  fourth  root  is  one  of  the  four  equal 
factors;  and  so  on  for  the  higher  roots. 

78.  Radical  sign  and  index  of  root. — To  indicate  a  root,  we 
use  the  sign  \/,  which  is  called  the  radical  sign.     A  small 
figure,  called  the  index  of  the  root,  is  placed  in  the  opening  of 
the  radical  sign  to  show  what  root  is  to  be  taken.     Thus,  -y/64 
indicates  the  cube  root  of  64.     The  small  3  is  the  index  of  the 
root. 

Since  the  square  root  is  the  most  frequently  written  root, 
the  index  2  is  omitted.  Thus,  the  square  root  of  625  is 
written  \/625  and  not  -v/625.  Higher  roots  are  indicated  as 
A/243,  \/l28. 

Evolution  is  the  process  of  finding  a  root  of  a  given  number. 

79.  Square  root— The  numbers  1,  4,  9,  16,  25,  36,  49,  64, 
81,  which  are  the  squares  of  the  numbers  1,  2,  3,  4,  5,  6,  7,  8,  9, 
respectively,    should   be   carefully  remembered.     It   will   be 
noticed  that  these  are  the  only  whole  numbers  less  than  100 
of  which  we  can  find  the  square  roots.     Such  numbers  as 
these  are  called  perfect  squares.     As  we  pass  to  numbers 
above  100,  the  perfect  squares  become  still  more  scarce. 

The  square  root  of  49  is  7,  but  the  square  root  of  56  cannot 


327 
284 


KM)  PRACTICAL  MATHEMATICS 

be  expressed  as  a  whole  number,  nor  can  it  be  expressed  as  a 
decimal  exactly.  We  can  find  it  to  any  desired  number  of 
decimal  places,  and  so  as  accurately  as  we  wish.  It  remains  to 
devise  a  method  by  which  this  may  be  done. 

The  practical  man  who  wishes  to  find  the  square  root  of  a 
number  does  not  care  greatly  why  he  goes  through  a  certain 
process;  but  it  is  very  important  to  him  that  he  shall  be  able 
to  find  the  root  quickly  and  accurately.  In  what  follows  then 
the  attempt  is  made  to  tell  in  as  simple  a  manner  as  passible 
how  to  find  the  root. 

80.  Process  for  the   square  root  of  a  perfect  square.— 

Example  1.     Find  \/522729. 

Explanation.  Process. 

First,  separate  the  number  52'27'29  (723  Ans. 

into   periods   of  two    figures  49 

each,  beginning  at  the  right,     142 
and  placing  a  mark  between 
them.    The  number  of  periods     1443 
thus  formed  is  equal  to  the 
number  of  figures  in  the  root. 

Find  the  largest  perfect  square  which  is  equal  to,  or  less 
than,  the  left-hand  period,  52.  This  perfect  square  is  49. 
Write  it  under  52;  and  put  its  square  root,  7,  to  the  right  as 
the  first  figure  of  the  root.  Now  subtract  49  from  52  and. 
bring  down  the  next  period,  27,  and  unite  with  the  remainder 
3,  thus  obtaining  327. 

Take  twice  7,  the  first  figure  of  the  root,  and  write  it  to  the 
left  of  327.  Find  how  many  times  this,  14,  is  contained  in  32, 
which  is  2,  for  the  second  figure  of  the  root.  Place  this  figure 
2  in  the  root,  and  also  to  the  right  of  14,  making  142.  Now 
multiply  142  by  2,  and  write  the  product,  284,  under  327. 
Subtract  284  from  327  and  bring  down  and  unite  the  next 
period,  29,  with  the  remainder,  43,  thus  obtaining  4329. 

In  the  above  work  327  is  called  the  first  remainder;  14, 
the  trial  divisor;  142,  the  true  divisor;  and  4329,  the  second 
remainder. 

Next  multiply  72  by  2,  and  write  it  at  the  left  of  4329  as  the 
second  trial  divisor.  Find  how  many  times  144  is  contained 


4ML*) 
4329 


20816 
20816 


POWERS  AND  ROOTS  101 

in  432,  which  is  3,  for  the  third  figure  of  the  root.  Place  this 
figure,  3,  in  the  root  and  also  at  the  right  of  144,  making 
1443,  the  second  true  divisor.  Multiply  1443  by  3  and  write 
the  product  under  4329.  This  gives  no  remainder.  There- 
fore, 723  is  the  exact  square  root  of  522,729,  that  is,  723X723 
-522,729. 
Example  2.  Find  \/67808T6. 

Explanation.  Process. 

First,   separate  the  6'78'08'16  (2604  Ans. 

number  into  periods  of  4 

two  figures  each   as  in       45    1273 
example  1.     As  before,  i276 

we   find   the   greatest       5204 
square,    4,   in    the  left- 
hand    period,    write  it 

under  6  and  put  the  square  root,  2,  of  this  square  for  the  first 
figure  of  the  root.  Subtract  the  square,  4,  from  6,  and  bring 
down  the  next  period,  78,  and  unite  it  with  the  2,  making  the 
first  remainder,  278. 

Take  twice  2  for  a  trial  divisor.  .  Find  how  many  times  it  is 
contained  in  the  first  remainder,  excepting  the  right-hand 
figure ;  that  is,  find  how  many  times  4  is  contained  in  27. 
The  number  is  6,  which  write  as  the  second  figure  of  the  root, 
and  also  at  the  right  of  the  trial  divisor.  This  makes  46 
the  true  divisor.  Multiply  the  true  divisor  by  6,  and  subtract 
the  product,  276,  from  the  first  remainder.  Bring  down  and 
unite  the  next  period  to  the  difference,  making  the  second 
remainder,  208. 

Multiply  the  root  already  found  by  2,  and  get  the  second 
trial  divisor,  52.  Find  how  many  times  this  is  contained  in 
20,  which  gives  0  for  the  next  figure  of  the  root.  Place  this  0 
in  the  root  and  also  to  the  right  of  52,  making  520,  the  second 
true  divisor.  Now,  since  the  0  written  in  the  root  is  the 
multiplier,  nothing  is  gained  by  multiplying  the  true  divisor 
by  it,  and  subtracting  from  208.  This  part  of  the  process  is 
omitted,  and  the  next  period,  16,  is  united  with  208,  making 
20,816,  the  third  remainder. 

The  third  trial  divisor  is  twice  the  root,  260,  which  gives 


102  PRACTICAL  MATHEMATICS 

520.  This  is  contained  in  2081,  4  times.  Place  the  4  as  the 
next  figure  of  the  root,  and  also  to  the  right  of  520,  making 
5204,  the  third  true  divisor.  Multiply  this  by  4  and  subtract 
from  the  third  remainder.  As  the  remainder  is  zero,  2604  is 
the  exact  square  root  of  6,780,816. 

81.  Square  root  of  a  number  containing  a  decimal.- 


Example.     Find  V 665. 1241.  Process. 

Here  the  division  into  periods  6'65.'12'41  (25.79  Ans. 

is  made  by  beginning  at  the 
decimal  point  and  going  in  both 
directions.  The  rest  of  the 
work  is  the  same  as  in  examples 


265 
225 


507 


4012 
3549 


1  and  2,  Art.  80.  5149T634T 

46341 

The  student  should  note  that  the  second  trial  divisor,  50, 
is  contained  8  times  in  the  first  three  figures  of  the  second 
remainder,  4012.  However,  if  8  were  used  as  the  root,  it 
would  give  a  number  larger  than  4012  when  the  true  divisor 
was  multiplied  by  it.  The  relations  noted  here  should  help 
to  make  clear  why  we  give  to  the  trial  divisor  its  name. 

The  decimal  point  in  the  root  is  so  placed  that  there  are 
as  many  whole  number  figures  in  the  root  as  there  are  whole 
number  periods  in  the  number  of  which  the  root  is  extracted. 
The  position  of  the  decimal  point  can  also  be  determined  so 
that  there  will  be  as  many  decimal  places  in  the  root  as  there 
are  decimal  periods  in  the  number  of  which  the  root  is  being 
extracted. 

If  the  decimal  part  of  the  number  consists  of  an  odd  number 
of  figures  a  cipher  is  annexed  to  make  a  full  period  at  the  right . 

Thus,  in  pointing  off  53.76542  into  periods  it  is  53'.76'54'20. 

82.  Roots  not  exact. — Most  numbers  are  not  perfect  squares, 
but  the  roots  may  be  found  to  any  desired  number  of  decimal 
places.  When  extracting  the  root  of  a  number  not  a  perfect 
square,  one  must  determine  how  many  decimal  places  he 
wishes  in  the  answer,  and  then  annex  ciphers  to  the  right  of  the 
number  till  there  are  as  many  decimal  periods  as  there  are 
to  be  decimal  places  in  the  root.  The  root  is  then  extracted 


POWERS  AND  ROOTS  103 

in  the  usual  manner.     We  stop  when  the  desired  number  of 
figures  is  found  in  the  root. 

Example.     Find  \/27  to  three  decimal  places  in  the  root. 

Explanation.  Process. 

Since  three  decimal  places  27.'00'00'00(5.196  Ans. 

are  required  in  the  root,  annex 


three  periods  of  ciphers  to  the 
right  of  27.    These  are  the  deci- 


1029 


9900 
9261 


mal  periods.     Extract  the  root 

as  before.     Place  the  decimal       ]Q386     53900 

point  in  the  root  as  in  example  62316 

of  Art.  81.     It  will  be  noticed  L584~ 

that    there    is    a    remainder; 

this  is  disregarded  as  it  affects  the  next  figures  only,  that  is, 

the  fourth  and  following  figures  in  the  decimal  part  of  the  root. 

83.  Root  of  a  common  fraction.  —  If  the  numerator  and 
the  denominator  of  the  fraction  are  each  a  perfect  square, 
find  the  square  root  of  each  separately. 

Example  1.     Find 


The  V  144  =  12,  and  \/625  =  25. 

Hence  V'iHfr£  =  H>      Ans. 

If  the  numerator  and  denominator  are  not  each  a  perfect 
square,  reduce  the  fraction  to  a  decimal  and  then  extract 
the  square  root  as  in  Art.  81. 

Example  2.     Find  \/f  • 

Reducing  to  a  decimal,  |-  =  0.28571428  •  •  •  . 
V0.28571428  =  0.5345. 

Hence,  \/f  =  0.5345  to  four  decimal  places. 

It  is  worth  noting  here  that  the  square  root  of  -f  may  be 
found  by  extracting  the  square  root  of  both  numerator  and 
denominator,  and  then  dividing  the  square  root  of  the  num- 
erator by  the  square  root  of  he  denominator.  This  process 
would  require  two  extractions  of  roots  and  one  long  division, 
and  so  make  the  work  about  three  times  what  it  is  if  the 
fraction  is  first  reduced  to  a  decimal  and  then  the  root 
extracted. 

84.  Short  methods.  —  Partly  division.  If  it  is  required  to 
extract  the  square  root  of  a  number  to,  say,  five  decimal 


104 


PRACTICAL  MATHEMATICS 


places,  making,  say,  seven  figures  in  the  root,  the  work  may 
be  shortened  by  extracting  the  root  in  the  usual  way  till  four 
figures  are  obtained,  and  then  dividing  the  last  remainder 
found  by  the  corresponding  trial  divisor  to  obtain  the  last 
three  figures  of  the  root.  In  general,  extract  root  till  more 
than  half  the  required  number  of  figures  are  found,  and  then 
for  the  other  figures  of  the  root  divide  the  remainder  by  the 
corresponding  trial  divisor. 

Process. 
1 78.  WOOWOO(  13.34 166    Ans. 


Example  1.  Find 
\/178  to  five  decimal 
places. 


23 


263 


78 
69_ 
900 

789 


2664 


11100 
10656 


2668) 


The  process  may  be 
contracted  still 
further  by  using  con- 
tracted division  when 
dividing. 


444000(166 
2668 
17720 
16008 


17120 

16008 

1112 


Method  by  factoring.  When  the  number  of  which  the  square 
root  is  to  be  extracted  can  be  factored  into  two  factors,  one 
of  which  is  a  perfect  square  and  the  other  the  number  2,  3,  5, 
6,  or  7,  a  very  useful  short  method  may  be  obtained.  For 
this  purpose  it  is  necessary  first  to  have  found  the  following 
square  roots: 

A/2  =  1.4142,  A/3  =  1.73205,  A/5  =  2.23607, 

A/6  =  2.4494,  \/7  =  2.6457. 

Of  these  the  most  useful  are  the  roots  of  2  and  3. 

Example  2.     Find  the  A/32. 
32=  16X2,  so  we  may  write 
A/32  =\/T6X  A/2  =  4  XI. 4 142  =  5.6568.  Ans. 

Example  3.     Find  \/125. 

A/1 25  =  A/25  X  A/5  =  5X2.236  =  11.180.  Ans. 


POWERS  AND  ROOTS  105 

86.  Rule  for  square  root. — After  carefully  following  through 
the  solutions  of  the  preceding  examples,  the  following  rule 
should  be  understood: 

RULE.  (1)  Begin  at  the  decimal  point  and  point  off  the  whole 
number  part  and  the  decimal  part  into  periods  of  two  figures  each. 
If  there  is  an  odd  number  of  figures  in  the  whole  number  part, 
the  left-hand  period  will  have  only  one  figure.  If  there  is  an 
odd  number  of  figures  in  the  decimal  part,  annex  a  cipher  so  that 
the  right-hand  period  shall  contain  two  figures. 

(2)  Find  the  greatest  square  in  the  left-hand  period  and  place 
it  under  that  period.     The  square  root  of  this  greatest  square  is 
the  first  figure  of  the  required  root.     Subtract  the  greatest  square 
from  the  left-hand  period  and  bring  down  and  unite  with  the 
remainder   the   next  period  of  the  number.     This  is  the  first 
remainder. 

(3)  Take  twice  the  root  already  found  for  a  trial  divisor,  which 
write  at  the  left  of  the  remainder.     Find  how  many  times  this 
trial  divisor  is  contained  in  the  remainder  omitting  the  right-hand 
figure.     This  gives  the  next  figure  of  the  root,  which  place  in  the 
root  and  also  at  the  right  of  the  trial  divisor,  forming  the  true 
divisor.     Multiply  the  true  divisor  by  the  figure  last  placed  in  the 
root  and  write  the  product  under  the  remainder.     Subtract  and 
bring  down  and  unite  the  next  period  in  the  number.     This  pro- 
cess is  repeated  for  each  figure  of  the  root. 

(4)  //  at  any  time  the  trial  divisor  will  not  be  contained  in 
the  corresponding  remainder,  place  a  cipher  in  the  root  and  at 
the  right  of  the  trial  divisor,  bring  down  another  period,  and 
continue  as  before. 

(5)  Point  off  in  the  root  as  many  decimal  figures  as  there  are 
decimal  periods  in  the  number  of  which  the  root  is  extracted. 

86,  Cube  root. — The  extraction  of  cube  root  is  so  seldom 
used  that  it  is  thought  best  to  omit  the  usual  consideration  of 
it.  It  is  found  in  a  very  simple  manner  by  the  use  of  loga- 
rithms, by  which  means  any  one  root  is  as  easily  found  as 
another.  (See  Art.  313.) 

EXERCISES  27 

Find  the  square  root  of  the  following : 

1.  516,961.  Ans.  719. 

2.  23,804,641.  Ans.  4879. 


1 1  If, 


I'UACTH'AL  MA  THEM  A  TICS 


3.  0.3364. 

4.  0.120409. 
6.  1159.4025. 

6.  2  to  four  decimals. 

7.  786,432  to  two  decimals. 

8.  7,326,456  to  two  decimals. 

9.  3  to  five  decimal  places. 

10.  5  to  three  decimal  places. 

11.  6  to  four  decimal  places. 

12.  7  to  five  decimal  places. 

13.  Wr 

14.  27  -f- 156.25  to  four  decimals. 


Ans.  0.58. 

Ana.  0.347. 

Ann.  34.05. 

Ans.   1.4142. 

Ans.  886.81. 

,4n*.  2706.74. 

Ans.  1.73205. 

Ans.  2.236. 

Ans.  2.4495. 

Ans.  2.64575. 

Ans.  TV 

Ans.  0.4157. 


Suggestion.     First  perform  the  division,  and  then  extract  the  root  of 
the  quotient. 

15.  I  to  four  decimal  places.  Ans.  0.8819. 

In  each  of  the  exercises  from  16  to  23,  carry  the  root  to  five  decimal 
places : 

16.  143.  Ans.  11.95826         20.  287.  Ans.  16.94107. 

17.  164.  Ans.  12.80624.         21.  396.  Ana.  19.89975. 

18.  92.  Ans.  9.59166.         22.  416.  Ans.  20.39608. 

19.  278.  Ans.  16.67333.         23.  539.  Ans.  23.21637. 
24.  Find  the  square  roots  of  the  following  by  short  methods:  (a)  28, 

(6)  72,  (c)  288,  (d)  75,  (e)  147,  (/)  192,  (g)  432. 

Ans.  (a)  5.2915,  (6)  8.4852,  (c)  16.971,  (d)  8.6603,  (e)  12.1244, 

if)  13.8564,  (g)  20.7846. 


Fio.  22. — Similar  figures. 


87.  Similar   figures. — The   following   principles   are   useful 
in  solving  many  problems: 

(1)  The  areas  of  similar  figures  are  in  the  name  ratio  a,s  the 
squares  of  their  like  dimensions. 

(2)  The  volumes  of  similar  solids  are  in  the  same  ratio  as 
the  cubes  of  their  like  dimensions. 

Similar  figures  are  such  as  have  the  same  shape. 
In  Fig.  22  the  following  pairs  are  similar:  (a)  and  (6);  (c) 
and  (d);  (e)  and  (/);  (g)  and  (h). 


POWERS  AND  ROOTS  107 

EXERCISES  28 

1.  If  the  diameter  of  (a)  is  6  in.  and  of  (b)  is  4  in.,  how  many  times  as 
large  as  (b)  is  (a)  ? 

Solution.     Area  of  (a):  area  of  (b)  =62:42=36:  16  =2£-  Ans. 

2.  Find  the  ratio  of  areas  of  (e)  to  (/),  if  the  shorter  side  of  (e)  is  9  ft. 
and  of  (/)  5  ft,  Ans.  3.24. 

3.  If  a  round  steel  rod  |  in.  in  diameter,  hanging  vertically,  will  sup- 
port 12,000  lb.,  what  will  a  rod  |  in.  in  diameter  support? 

Ans.  36,750  lb. 

4.  Given  that  the  electrical  resistance  is  inversely  in  the  same  ratio  as 
the  areas  of  the  cross  sections  of  the  conductors  of  the  same  material; 
find  the  ratio  of  the  resistances  of  two  copper  wires  of  diameters  |  in. 
and  3  in.  respectively.  Ans.  64:9. 

6.  Two  steam  boilers  of  the  same  shape  are  respectively  12  ft.  and  18 
ft.  long.  Find  the  ratio  of  their  surfaces.  Ans.  4:9. 

6.  How  many  times  as  much  gold  leaf  will  it  take  to  cover  a  ball  10 
in.  in  diameter  than  to  cover  a  ball  6  in.  in  diameter?  Ans.  2  f  • 

7.  Two  balls  of  steel  are  respectively  7  in.  and  15  in.  in  diameter. 
The  second  is  how  many  times  as  heavy  as  the  first?         Ans.   9.84  —  . 

8.  Which  is  the  cheaper,  oranges  2J  in.  in  diameter  at  30  cents  a  dozen 
or  3-in.  oranges  at  40  cents  a  dozen?     What  should  the  larger  ones  sell 
at  to  give  the  same  value  for  the  money  as  the  smaller  at  30  cents  a 
dozen?  Ans.  The  3-in.  oranges;  52  cents  a  dozen  nearly. 

Suggestion.  The  price  the  3-in.  oranges  should  sell  at  is  given  by  the 
proportion:  (2|)3:33=30:x. 

9.  Two  balls  of  the  same  material  are  10  in.  and  3  in.  in  diameter 
respectively.     If  the  smaller  ball  weighs  9  lb.  what  is  the  weight  of  the 
larger?  Ans.  333^  lb. 

10.  The  formula  V  =  \/2gh  gives  the  velocity  V  in  feet  per  second  a 
body  will  have  after  falling  from  a  height  h.     Find  the  value  of  V  for  a 
stone  that  has  fallen  400  ft.     In  the  formula  0=32.2. 

Ans.  160.5  ft.  nearly. 

Suggestion.  As  in  this  exercise,  the  evaluation  of  a  formula  often  re- 
quires the  extraction  of  a  square  root.  The  numbers  that  the  letters 
stand  for  are  put  in  place  of  the  letters  and  we  have 

V  =  A/2X32.2X400  =  V25760  =  160.5  -  . 

11.  The  effective  area  of  a  chimney  is  given  by  the  formula 


where  £'  =  the  effective  area,  and  A=the  actual  area  of  the  flue.     Find 
the  effective  area  if  A  =86  sq.  in.     If  A  =3.14  sq.  ft. 

Ans.  85.44  sq.  in.;  3.03  sq.  ft. 

12.  When  the  pressure  of  water  at  the  place  of  discharge  is  known,  the 
rate  of  flow  is  given  by  the  formula 


108  PRACTICAL  MATHEMATICS 

where  F=velocity  of  discharge  in  feet  per  second,  and  P=pres8ure  in 
pounds  per  square  inch  at  the  place  of  discharge.  Find  the  rate  of  dis- 
charge if  the  pressure  as  given  by  a  pressure  gage  is  50  Ib.  per  square 
inch.  Ans.  85.98  ft.  per  second. 

IS.  As  in  the  last,  find  the  velocity  of  discharge  if  the  pressure  is  200 
Ib.  per  square  inch.     Compare  the  result  with  that  of  the  preceding. 

Ans.  171.97  ft.  per  second 


PART  TWO 
GEOMETRY 

CHAPTER  X 

PLANE  SURFACES.    LINES  AND  ANGLES 

88.  In  this  and  the  following  chapters  are  discussed  some 
of  the  facts  established  in  geometry,  and  some  of  their  appli- 
cations to  practical  problems.     The  endeavor  is  to  illustrate 
and  make  clear  the  principles  and  thus  lay  a  broad  foundation, 
rather  than   to  follow  narrow  special   lines.     Many  special 
problems,   however,   are   given.     From   these  the  individual 
student  can  select  those  that  are  suited  to  his  needs. 

There  are  many  terms  which,  although  quite  familiar  to 
the  student,  are  used  in  geometry  with  such  exactness  as  to 
require  a  careful  definition  or  explanation.  Point,  line, 
angle,  surface,  and  solid  are  such  terms.  Like  all  simple 
terms,  such  as  number,  space,  and  time,  they  are  difficult 
to  define;  but  it  is  hoped  the  explanations  given  will  lead  to  a 
reasonable  understanding  of  them. 

89.  Definitions. — A    material    body,    as,    for    example,    a 
block  of  wood  or  an  apple,  occupies  a  definite  portion  of  space. 

In  geometry  no  attention  is  given  to  the  substance  of  which 
the  body  is  composed.  It  may  be  iron,  stone,  wood,  or  air, 
or  it  may  be  a  vacuum.  Geometry  only  considers  the  space 
occupied  by  the  substance.  This  space  is  called  a  geometric 
solid  or  simply  a  solid. 

If  one  thinks  of  a  brick,  and  then  considers  the  brick  removed  and 
thinks  of  the  space  that  the  brick  occupied,  he  has  an  illustration  of  a 
geometric  solid. 

A  solid  has  length,  breadth,  and  thickness. 
A  boundary  face  of  a  solid  is  called  a  surface. 
A  surface  has  length  and  breadth  but  no  thickness. 

109 


110 


I'K. \  ('  TIC  A  L  MA  THEM  A  TICK 


The  boundary  of  a  surface,  or  that  which  separates  one 
part  of  a  surface  from  an  adjoining  part,  is  called  a  line. 

A  line  has  length  only. 

That  which  separates  one  part  of  a  line  from  an  adjoining 
part  is  called  a  point. 

A  point  has  neither  length,  breadth,  nor  thickness.  It  has 
position  only. 

A  point  is  read  by  naming  the  letter  placed  upon  it.  A  line 
is  read  by  naming  the  letters  placed  at  its  ends,  or  by  naming 


B      D 


Fio.  23. 

the  single  letter  placed  upon  it.  Capital  letters  are  usually 
used  at  the  ends  of  a  line,  while  a  small  letter  is  placed  upon 
a  line.  In  Fig.  23(a),  the  line  is  read  "  the  line  AB"  or  simply 
"the  line  a." 

A  straight  line  is  a  line  having  the  same  direction  through- 
out its  whole  extent.  See  Fig.  23(a). 

A  curved  line  is  a  line  that  is  continually  changing  in  direc- 
tion. See  Fig.  23(6). 

A  broken  line  is  a  line  made  up  of  connected  straight  lines. 
See  Fig.  23  (c). 


Plane  Surface 


Curved  Surface 
Fin.  24. 


Parallel  Lin*. 


If  a  surface  is  such  that  any  two  points  in  it  can  be  con- 
nected by  a  straight  line  lying  wholly  in  the  surface,  it  is 
called  a  plane  surface  or  simply  a  plane. 

A  carpenter  determines  whether  or  not  the  surface  of  a 


PLANE  SURFACES.     LINES  AND  ANGLES 


board  is  a  plane  by  laying  the  edge  of  his  square  or  other 
straightedge  on  the  surface  in  different  positions,  and  observ- 
ing if  the  straightedge  touches  the  surface  at  all  points. 

A  curved  surface  is  a  surface  no  part  of  which  is  a  plane 
surface.  Thus,  the  surface  of  a  circular  pipe  and  the  surface 
of  a  ball  are  curved  surfaces. 

Parallel  lines  are  lines  in  the  same  plane  and  everywhere 
the  same  distance  apart. 

In  Fig.  24  are  shown  pairs  of  parallel  lines. 

90.  Angles. — Two  straight  lines  which  meet  at  a  point 
form  an  angle.  The  idea  of  what  an  angle  is,  being  a  simple 
one,  is  hard  to  define.  One  should  guard 
against  thinking  of  the  point  where  the 
two  lines  meet  as  the  angle.  This  point 
is  called  the  vertex  of  the  angle. 

The   two   lines   are   called   the  sides  of 
the   angle.      The   difference  in  the  direc- 
tions of  the  two  lines  forming  the  angle  is 
the  magnitude  of  the  angle.     For  a  further  discussion  of  an 
angle  see  Art.  317. 

An  angle  is  read  by  naming  the  letter  at  the  vertex,  or  by 
naming  the  letters  at  the  vertex  and  at  the  ends  of  the  sides. 


FIG.  25. 


s 

'I 

« 

B 

0) 

3 

,-,       Horizontal 

D     Line 

1  11111  i  llllJ_ii_LLJJJ_LJ_l_[_r 

n                            P 

FIG.  26. 

When  read  in  the  latter  way,  the  letter  at  the  vertex  must 
always  come  between  the  other  two. 

Thus,  the  angle  in  Fig.  25  is  read  "the  angle  b,"  "the  angle  ABC,"  or 
''the  angle  at  B." 

If  one  straight  line  meets  another  so  as  to  form  equal  angles, 
the  angles  are  right  angles,  and  the  lines  are  perpendicular 
to  each  other. 

In  Fig.  26  (a\  lines  AB  and  CD  are  perpendicular  to  each  other. 


112 


PRACTICAL  MA  THEM  A  TICK 


A  vertical  line  or  a  plumb  line  is  the  line  along  which  a 
string  hangs  when  suspended  at  one  end  and  weighted  at  the 
other. 

A  horizontal  line  is  a  line  that  is  perpendicular  to  a  vertical 
line.  Fig.  26(6). 

If  a  right  angle  is  divided  into  90  equal  parts,  each  part  is 
called  a  degree.  It  is  usually  written  1°. 

An  acute  angle  is  an  angle  that  is  less  than  a  right  angle. 
An  obtuse  angle  is  an  angle  that  is  greater  than  a  right  angle 
and  less  than  two  right  angles.  See  Fig.  26(c). 


D    C 


E  B 

Complementary 
Ansrles 


EB 

Supplementary 
Angles 


FIG.  27 


Two  angles  whose  sum  is  one  right  angle,  or  90°,  are  called 
complementary  angles,  and  either  one  is  said  to  be  the  com- 
plement of  the  other.  Two  angles  whose  sum  is  two  right 
angles,  or  180°,  are  called  supplementary  angles,  and  either 
one  is  said  to  be  the  supplement  of  the  other. 

SURFACES 

91.  Polygons. — A  polygon  is  a  plane  surface  bounded  by 
any  number  of  straight  lines.  Any  one  of  these  lines  is  called 


D 


B 


FUJ.  L'S. 


:i  side.  The  point  where  two  sides  meet  is  called  a  vertex. 
The  distance  measured  around  the  polygon,  or  the  sum  of  the 
lengths  of  the  sides,  is  called  the  perimeter  of  the  polygon. 


PLANE  SURFACES.     LINES  AND  ANGLES 


113 


A  triangle  is  a  polygon  having  three  sides. 

A  quadrilateral  is  a  polygon  having  four  sides. 

A  pentagon  is  a  polygon  having  five  sides. 

A  hexagon  is  a  polygon  having  six  sides. 

An  octagon  is  a  polygon  having  eight  sides. 

A  regular  polygon  is  one  whose  sides  are  all  equal  and  whose 
angles  are  all  equal. 

A  diagonal  is  a  line  joining  any  two  vertices  not  adjacent 
in  a  polygon. 

92.  Concerning  triangles. — A  line  drawn  from  any  vertex 
of  a  triangle  perpendicular  to  the  opposite  side  and  ending  in 
it  is  called  an  altitude  of  the  triangle.  Since  a  triangle  has 
three  vertices,  each  triangle  has  three  altitudes.  The  altitude 


^ 


D 

FIG.  29. 


FIG.  30. 


may  meet  the  opposite  side,  as  CF  in  triangle  ABC,  Fig.  29; 
or  the  opposite  side  may  have  to  be  extended  to  meet  it,  as 
AD  and  BE,  Fig.  29. 

A  line  drawn  from  any  vertex  of  a  triangle  to  the  center  of 
the  opposite  side  is  called  a  median.  It  is  evident  that  in  any 
triangle  there  are  three  medians. 

In  Fig.  30,  AD  is  a  median. 

A  line  drawn  through  the  vertex  of  an  angle  and  dividing 
the  angle  into  two  equal  parts  is  called  the  bisector  of  the 
angle.  The  bisector  of  an  angle  of  a  triangle  is  often  taken  as 
the  length  of  the  bisector  of  an  angle  of  the  triangle  from  the 
vertex  to  the  opposite  side. 

BE  in  Fig.  30  is  the  bisector  of  the  angle  ABC  of  the  triangle. 


1 1 4  I'K  A  CTICAL  MA  THEM  A  TICK 

It  is  evident  that  there  are  three  bisectors  of  the  angles  in 
any  triangle. 

93.  Concerning  quadrilaterals. — A  parallelogram  is  a  quad- 
rilateral whose  opposite  sides  are  parallel.  See  Fig.  31  (a). 

A  rectangle  is  a  parallelogram  whose  angles  are  right  angles. 
See  Fig.  31(6). 

A  square  is  a  rectangle  whose  sides  are  all  equal.  See 
Fig.  31(c). 


FIG.  31. 

A  trapezoid  is  a  quadrilateral  with  only  two  sides  parallel. 
The  parallel  sides  are  called  the  bases.  The  altitude  is 
the  distance  between  the  two  bases. 

Fig.  31  (d)  is  a  trapezoid;  AB  and  DC  are  the  bases,  and  EF  is  the 
altitude. 

The  forms  just  discussed  are  very  important,  as  any  figure 
bounded  by  straight  lines  may  be  thought  of  as  composed 
of  rectangles  and  triangles. 

EXERCISES  29 

In  the  following  exercises  use  a  ruler  and  a  hard  lead  pencil.  Letter 
all  figures. 

1.  Draw  two  curved  lines.     Two  broken  lines. 

2.  Draw  several  parallel  lines. 

3.  Draw  a  right  angle.     An  acute  angle.     An  obtuse  angle. 

4.  Draw  perpendicular  lines.     If  two  lines  are  perpendicular  to  each 
other  is  one  of  them  vertical?     Illustrate  by  a  drawing. 

6.  Draw  vertical  and  horizontal  lines.  Is  a  vertical  line  always 
perpendicular  (o  a  horizontal  line? 

6.  Estimate  the  size  as  nearly  as  you  can  and  draw  an  angle  of  45°. 
Of    30°.     Of   60°.     Of    120°.     Of    135°.     Of    180°.     Which    are   acute 
angles?     Which  obtuse  angles? 

7.  Draw  two  complementary  angles.     Two  supplementary  angles. 

8.  Draw  a  triangle.     A  quadrilateral.     A  pentagon.     A  hexagon.     An 
octagon.     A  regular  hexagon. 

9.  How  many  diagonals  have  each  of  the  polygons  of  exercise  8? 

10.  What  are  the  vertices  of  each  polygon  of  exercise  8?     What  are 
the  perimeters? 


PLANE  SURFACES.     LINES  AND  ANGLES 


115 


B 


11.  Draw  a  triangle  having  all  its  angles  acute,  and  draw  its  three 
altitudes. 

12.  Draw  a  triangle  having  all  its  angles  acute,  and  draw  its  three 
medians.     Draw  the  three  bisectors  of  its  angles. 

13.  Draw  triangles  each  having  one  obtuse  angle,  and  follow  the  direc- 
tions of  exercises  11  and  12. 

14.  Draw  a  rectangle.     A  square.     A  parallelogram.     A  trapezoid. 
A  quadrilateral  that  is  not  any  of  these.     Draw  their  altitudes. 

15.  Name  objects  in  nature,  or  objects  made  by  man  that  are  of  the 
forms  asked  for  in  the  preceding  exercises. 

AREAS  OF  POLYGONS 

94.  The  rectangle. — How  to  find  the  area  of  a  rectangle 
is  illustrated  in  Fig.  32.  Suppose  that  this  represents  a 
rectangle  whose  length  AD  is  5  ft., 
and  width  AB  is  4  ft.  The  rectangle 
is  divided  into  small  squares  1  ft.  on 
a  side,  and  so  each  represents  1  sq. 
ft.  Since  there  are  4  rows  of  squares 
each  containing  5  sq.  ft.,  there  are 
4X5  sq.  ft.  =20  sq.  ft.  in  the  rectan- 
gle. What  is  said  will  also  be  true  if  . 
the  lengths  of  the  sides  are  frac- 
tional. This  leads  to  the  following: 

RULE.     The  area  of  a  rectangle  is  equal  to  the  product  of 
its  length  and  its  width. 

Remark.  The  length  and  the  width  of  the  rectangle  must 
be  in  the  same  unit  before  taking  their  product.  The  product 
is  then  square  units  of  the  same  kind  as  the  linear  units. 
( Thus,  if  the  unit  of  length  is  the  foot, 
the  product  will  be  square  feet. 

95.  The  parallelogram. — A  paral- 
lelogram and  a  rectangle,  each  hav- 
ing the  same  base  and  altitude,  are 
equal  in  area.  This  is  illustrated  in 
Fig.  33.  A  BCD  is  the  rectangle  and 
ABEF  is  the  parallelogram.  The  altitude  BC  is  the  same  for 
each,  and  they  have  the  same  base,  AB.  Since  the  part 
BCE  of  the  parallelogram  may  be  cut  off  and  fitted  on  ADF, 
it  is  evident  that  the  parallelogram  is  just  equal  to  the  rec- 
tangle. Therefore,  we  have  the  following: 


D 


FIG.  32. 


E 


FIG.  33. 


1  1  (i  I'KA  CTICAL  MA  THEM  A  TICK 

RULE.  The  area  of  a  parallelogram  is  equal  to  the  product 
of  its  base  and  its  altitude. 

96.  Formulas.  —  A  rule  stated  in  letters  and  signs  is  called 
a  formula.    It  is  a  shorthand  way  of  stating  a  rule. 

If  A  is  used  as  an  abbreviation  for  area,  b  for  base,  and 
a  for  altitude,  the  rule  for  the  area  of  a  rectangle  or  a  paral- 
lelogram is  given  in  the  following  formula: 

[I]  A  =  ab. 

The  form  ab  means  altitude  times  base. 

Since  the  altitude  times  the  base  equals  the  area,  by  using 
well-known  principles  of  division  we  have  for  the  rectangle  or 
parallelogram  the  following: 

RULE.  (1)  The  altitude  equals  the  area  divided  by  the  base. 
(2)  The  base  equals  the  area  divided  by  the  altitude. 

These  rules  written  as  formulas  are  : 

[2]  a  =  A-^b, 
[3]  b  =  A^a. 

97.  The  triangle.  —  If  a  triangle  and  a  parallelogram  have 
the  same  base  and  have  their  altitudes  equal,  the  triangle  has 
half  the  area  of  the  parallelogram. 

This  is  illustrated  in  Fig.  34.     ABCD  is  the  parallelogram. 
Q  The  diagonal  BD  divides  it  into  two  tri- 
angles ABD  and  BCD,  which  are  equal. 
From  this  and  the  rule  for  the  area  of 
a  parallelogram,  it  is  clear  that  the  fol- 
lowing is  true  : 

RULE.       The  area  of  any  triangle  is 
equal  to  one-half  of  its  base  times  its  altitude. 

If  the  area  and  either  base  or  altitude  of  a  triangle  are  given, 
the  other  dimension  (altitude  or  base)  is  found  by  dividing 
twice  the  area  by  the  given  dimension. 

If  A  stands  for  the  area,  a  for  the  altitude,  and  6  for  the  base, 
we  have  these  formulas  for  the  triangle  : 


[4]  A  = 

[5]  a  =  2A-^-b, 

[6]  b=2A-a, 


PLANE  SURFACES.     LINES  AND  ANGLES  117 

EXERCISES  30 

1.  Compute  the  areas  of  the  following  figures  using  the  dimensions  as 


given. 


22  rd. 


FIG.  35. 


2.  If  the  sides  only  of  a  parallelogram  are  given  can  its  area  be  found? 

3.  Draw  two  triangles  and  find  their  areas  by  drawing  the  three  alti- 
tudes of  each  and  measuring  the  sides  and  altitudes. 

98.  Area  of  a  triangle  when  the  three  sides  only  are  given.  — 

If  a,  b,  and  c  stand  for  the  three  sides  of  a  triangle;  and  if  s 
stands  for  one-half  the  sum  of  a,  b,  and  c,  then  the  area  A 
of  the  triangle  is  given  by  the  formula: 
[7]  A  = 


a)  (s-b)  (s-c). 

This  formula  cannot  well  be  derived  here,  but  it  is  found  in 
geometry.  The  area  of  the  triangle  can  also  be  found  by 
constructing  it  to  scale,  as  explained  later.  The  altitude  can 
then  be  measured  and  the  area  be  found  by  taking  one-half 
the  product  of  the  base  and  the  altitude. 

Since  a  formula  is  a  rule  stated  in  symbols,  the  above  formula 
may  be  stated  as  the  following  rule  for  the  area  of  a  triangle 
when  the  three  sides  only  are  given  : 


118 


PRACTICAL  MA  THEM  A  TICS 


HULK.  Fiiul  half  the  sum  of  the  three  sides.  Subtract 
each  side  from  this  half  sum.  Take  the  continued  product  of  the 
half  fnim  and  the  three  differences.  The  square  root  of  this 
product  z.s  the  area  of  the  triangle. 

This  rule  can  be  illustrated  best  by  an  example. 

Example.  Find  the  area  of  a  triangle  with  sides  40  rd., 
28  rd.,  and  36  rd. 

Solution,     a  =  40,  6  =  28,  c  =  36. 

s  =  i(40+28+36)  =  52. 
s-a  =  52-40  =  12. 
s-6  =  52-28  =  24. 
s-c  =  52-36  =  16. 

A  =  \/52X  12X24X16  =  \/239,616  =  489.506. 
/.area  =  489. 506-  rd.2     Ans. 


With  very  ordinary  instruments  this  triangle  can  be  con- 
structed to  scale  and  measured,  and  the  area  found  to  within 
half  a  square  rod  of  the  computed  area. 

99.  Area  of  trapezoid. — A  diagonal  of  a  trapezoid  divides 
it  into  two  triangles  which  have  the  same  altitude,  and  have 

as  bases  the  two  bases  of  the  trape- 
zoid. Thus,  in  the  trapezoid  of  Fig. 
38,  the  diagonal  AD  divides  the 
trapezoid  into  two  triangles  A  CD 
and  ADE.  The  area  of  A  CD  =  $  of 
ACXa  and  area  of  ADE  =  \  of 
EDXa'.  But  a  =  o',  hence  the  sum 
of  the  areas  of  the  two  triangles  =  $ (AC +ED)X a. 

Now  the  area  of  the  trapezoid  can  evidently  be  found  by 
finding  the  sum  of  the  areas  of  the  two  triangles  into  which 


B 

Fir;. 


PLANE  SURFACES.     LINES  AND  ANGLES  119 

it  is  divided;  or  what  amounts  to  the  same  thing,  by  the  fol- 
lowing: 

RULE.  The  area  of  a  trapezoid  equals  one-half  the  sum  of 
the  two  bases  times  the  altitude. 

If  B  and  b  stand  for  the  two  bases  and  a  for  the  altitude  of 
the  trapezoid,  the  formula  is 

[8]  A=i(B+b)Xa. 

Example.  Find  the  area  of  a  trapezoid  whose  lower  base 
is  20  rd.,  upper  base  14  rd.,  and  altitude  9  rd. 

Solution.  By  formula  [8],  A  =  %(B-\-b)a.  Putting  the  num- 
bers of  the  example  in  place  of  the  letters  of  the  formula, 


.'.  area  =  153  sq.  rd.     Ans. 

EXERCISES  31 

1.  Find  the  parts  not  given  in  the  following  exercises  which  refer  to 
parallelograms: 

(1)  Base  22|  in.  altitude  19  in.  area  —          —  . 

(2)  Base  -       —  altitude  47  rd.  area  426  rd.2 

(3)  Base  33^  ft.  altitude  -       -  area  433f  ft.2 

(4)  Base  —       —  altitude  102§  in.  area  9367  in.2 

Ans.  (1)  427|  in.2;  (2)  9*37rd.;  (3)  13^  ft.;  (4)  9U{ft  in. 

2.  Find  the  number  of  acres  in  a  farm  160  rd.  long  and  80  rd.  wide. 

Ans.  80. 

3.  Find  the  number  of  square  feet  in  a  floor  16  ft.  8  in.  by  13  ft.  6  in. 

Ans.  225. 

4.  Find  the  number  of  square  meters  in  a  rectangle  77  m.  long  and 
5  Dm.  wide.  Ans.  3850. 

5.  A  box  6  in.  long,  4  in.  wide,  and  3  in.  deep  has  six  rectangular  faces. 
Find  the  area  of  the  surface  of  the  box.  Ans.  108  in.2 

6.  Find  the  area  of  a  triangle  whose  base  is  25  ft.  and  whose  altitude  is 
12  ft.  4  in.  Ans.   154£  ft.2 

7.  How  many  acres  are  there  in  a  triangular  lot  whose  base  is  432  ft. 
and  altitude  320  ft.?  Ans.  1.59-. 

8.  Find  the  number  of  hectares  in  a  triangular  field  whose  base  is 
196.8  m.  and  altitude  85  m.  Ans.  0.8364. 

9.  Find  the  base  of  a  triangle  whose  area  is  20  acres  and  altitude  80  rd. 

Ans.  80  rd. 

10.  A  rectangular  field  48  rd.  long  contains  9  acres.     Find  the  width. 

Ans.  30  rd. 

11.  If  the  perimeter  of  a  rectangle  is  96  ft.  and  the  length  is  three 
times  the  breadth,  find  the  area.  Ans.  432  ft.'- 


120  PRACTICAL  MATHEMATICS 

12.  A  rectangular  garden  56  ft.  long  and  40  ft.  wide  has  a  path  6  ft. 
wide  around  it.     Find  the  area  of  the  path.  Ans.  1296  ft.1 

13.  A  box  of  tin  sheets  for  roofing,  containing  112  sheets  14  in.  by 
20  in.,  will  cover  170  ft.1     What  per  cent  of  surface  covered  is  allowed 
for  joints  and  waste? 

Solution.     Without  allowing  for  joints  and   waste  each  box   would 

14X20X112 
cover-    ~i4l =  2175  sq.ft. 

217$  sq.  ft.  — 170  sq.  ft.  =  47J  sq.  ft.  =  allowance  for  joints  and  waste. 
47J  sq.  ft.  +170  sq.  ft.  =0.28+  =28  +  %. 

14.  How  many  bricks  each  9  in.  by  4J  in.  by  1}  in.  will  it  take  to  pave 
a  court  16  ft.  by  18  ft.,  if  bricks  are  laid  flat?     If  laid  on  edge? 

Ans.  flat  1024;  edge  2634. 

16.  How  many  paving  blocks  each  4  in.  by  4  in.  by  10  in.,  placed  on 
their  sides,  will  it  take  to  pave  an  alley  600  ft.  long  and  12  ft.  6  in.  wide? 

Ans.  27,000. 

16.  What  will  be  the  expense  of  painting  the  walls  and  ceiling  of  a 
room  12  ft.  6  in.  by  16  ft.  and  10  ft.  4  in.  high  at  15  cents  per  square 
yard?  Ans.  $13.15. 

17.  Find  the  cost  of  sodding  a  lawn  31  ft.  wide  and  52  ft.  long  at  18 

cents  per  square  yard. 

Ana.  $32.24. 

18.  Find  the  number  of 
square  feet  in  the  floor  of  the 
room  shown  in  Fig.  39. 

Ans.  277i  ft.1 

Suggestion.  Divide  into 
rectangles  and  trapezoids. 

JTIG    3Q_  19.  At    15  cents   per  square 

foot,   find   the  cost  of  building 

a  cement  walk  6  ft.  wide,  on  two  sides  of  a  corner  lot  33  ft.  by  100  ft. 

Ans.  $125.10. 

20.  Find  the  area  of  a  trapezoid  whose  bases  are  17  in.  and  11  in. 
respectively  and  whose  altitude  is  13  in.  Ans.  182  in.1 

21.  Find  the  area  of  a  triangle  whose  sides  are  13  in.,  15  in.,  and  21  in. 

Ans.  96.79-  in.* 

22.  Find  the  area  of  a  triangle  whose  sides  are  54  in.,  32  in.,  and  22  in. 

Ans.  0  in.1 

23.  Find  the  area  of  a  triangle  whose  base  is  27  in.  and  altitude  14  in. 

Ans.  189in.» 

24.  Find  the  area  of  a  board  14  ft.  long  and  18  in.  wide  at  one  end  and 
12  in.  at  the  other.  Ans.  17.5  ft.1 

26.  Find  the  cost  of  painting  both  sides  of  a  solid  board  fence  260  ft. 
long  and  6  ft.  high  at  60  cents  a  square.  How  many  gallons  of  paint 
will  it  take  for  two  coats  if  1  gallon  will  cover  250  sq.  ft.  two  coats? 
(1  square  =  100  sq.  ft.)  Ans.  $18.72;  12$  gal.  nearly. 

26.  How  much  did  it  cost  to  harvest  a  field  of  wheat  156  rd.  by  76  rd., 


PLANE  SURFACES.     LINES  AND  ANGLES 


121 


if  cutting  and  binding  cost  $1.50  per  acre,  setting  up  25  cents  an  acre, 
and  hauling  $1.25  an  acre?  Ans.  $222.30. 

27.  Find  the  area  in  acres  of  a  farm  which  is  represented  on  paper  as 
a  rectangle  3f  in.  by  102  in.  on  a  scale  of  •?$  in-  to  the  rod. 

Ans.  63  A. 

28.  Find  the  area  of  Fig.  40(o).  Ans.  23.592  in.2 

29.  Find  the  area  of  Fig.  40(6).  Ans.  6.02  in.2 

30.  Find  the  area  of  Fig.  40(c).  Ans.  8.625  in.2 


.—S.40 


FIG.   40. 


31.  Find  the  area  of  the  footing  for  a  column  with  a  load  of  168,000  Ib. 
if  the  safe  bearing  load  of  the  soil  is  4000  Ib.  per  square  foot. 

Ans.  42  sq.  ft. 

32.  How  many  square  yards  of  plastering  will  be  required  for  the  four 
side  walls  of  a  hall  90  ft.  long,  50  ft.  wide,  and  20  ft.  high,  with  4  doors 
5J  ft.  by  10  ft.,  14  windows  5  ft.  by  11  ft.,  and  a  baseboard  9  in.  high 
around  the  room?     Find  the  cost  at  40  cents  per  square  yard.     Find  the 
contractor's  profit  at  20%. 


LUMBER 


100.  Measuring  lumber. — Lumber  is  measured  in  board 
measure.  Timber  used  in  framework  is  counted  as  lumber. 
Lumber  and  timber  are  sold  by  the  1000  ft.  board  measure. 
This  is  sometimes  written  1000  ft.  B.M.,  but  more  often  it  is 
indicated  by  the  single  letter  M. 


122  PRACTICAL  MATHEMATICS 

One  board  foot  is  12  in.  square  and  1  in.  thick,  and  so  con- 
tains one-twelfth  of  a  cubic  foot.  The  number  of  board  feet 
in  a  stick  of  timber  is  the  number  of  cubic  feet  times  12. 
The  following  rule  may  be  used  to  find  the  number  of  board 
feet  in  a  stick  of  timber: 

RULE.  Take  the  product  of  the  end  dimensions  in  inches, 
divide  by  12,  and  multiply  the  quotient  by  the  length  in  feet. 

The  student  should  make  clear  to  himself  the  correctness 
of  this  rule. 

Example.  Find  the  number  of  board  feet  in  a  stick  of 
timber  6  in.  by  8  in.  and  14  ft.  long. 


Solution.  xl4  =  56ft.  B.M.  Ans. 


Lumber  less  than  1  in.  is  counted  as  if  1  in.  thick  in  buying 
and  selling.  In  widths  a  fraction  of  \  in.  or  more  is  counted 
as  1  in. 

Usually  lumber  is  cut  in  lengths  containing  an  even  number 
of  feet,  as  12,  14,  and  16  ft.  Longer  lengths  than  these 
are  usually  special,  but  classifications  vary  greatly.  There 
are  sixteen  or  more  associations  in  America  with  specifications 
governing  the  classification  of  lumber,  and  these  specifications 
differ  more  or  less. 

Timber  work  is  usually  paid  for  at  an  agreed  price  per  M, 
the  timber  to  be  measured  in  the  work. 

101.  Estimations.  —  There  are  various  rules  regarding  the 
estimating  of  the  amount  of  lumber  required  in  a  structure. 
In  general,  all  that  is  necessary  is  to  find  the  number  of  board 
feet  in  the  lumber  required  and  add  a  certain  per  cent  for 
waste  in  cutting,  matching,  etc.     Regardng  this,  the  student 
can  consult  a  handbook  specially  prepared  for  those  in  this 
line  of  work. 

102.  Shingles.  —  Shingles  are  16  in.  or  18  in.  in  length,  are 
counted  as  4  in.  wide,  and  put  up  in  bunches  of  250.     The  part 
of  the  shingle  that  is  exposed  when  laid  is  said  to  be  "laid  to 
the  weather."     The  part  so  exposed  varies  from  4  in.  to  6  in. 
So  a  single  shingle  covers  a  space  4  in.  wide  and  from  4  in.  to 
6  in.  long. 

In  laying  shingles,  the  estimating  is  often  made  by  the 
square,  an  area  10  ft.  by  10  ft.  or  containing  100  sq.  ft. 


PLANE  SURFACES.     LINES  AND  ANGLES  123 

In  stating  the  number  of  shingles,  give  the  number  so  that 
only  whole  bunches  will  be  required.  Thus,  do  not  give  a 
number  as  5550  but  as  5750. 

The  following  table  allows  for  waste  and  gives  the  number 
of  square  feet  covered  by  a  thousand  shingles,  and  also  the 
number  of  shingles  required  to  cover  a  square,  when  laid  at 
various  distances  to  the  weather. 

Inches  to  the  Area  covered  by  1000  No.  to  cover 

weather  shingles  a  square 

4   100  sq.  ft 1000 

4i 110  sq.  ft 910 

4) 120  sq.  ft 833 

5   133  sq.  ft 752 

5| 145  sq.  ft 690 

6  157  sq.  ft 637 


EXERCISES  32 

1.  Find  the  number  of  feet  of  lumber  it  will  take  to  build  a  tight  board 
fence  5|  ft.  high  and  70  ft.  long,  boards  1  in.  thick  and  nailed  at  top  and 
bottom  to  pieces  of  2  in.  by  4  in.  stuff.     (No  waste  allowed.) 

Ans.  478. 

2.  Find  cost  of  lumber  at  $32.00  per  M  to  build  a  walk  30  ft.  long  and 
8  ft.  wide;  plank  to  be  2  in.  thick  and  laid  crosswise  on  4  pieces  of  4  in. 
by  4  in.,  running  lengthwise.  Ans.  $20.48. 

3.  Find  the  amount  of  lumber  to  floor  a  room  30  ft.  by  40  ft.  with 
strips  3  in.  wide,  allowing  £  for  matching  and  15%  for  waste. 

4.  Find  how  many  shingles  it  will  take  to  shingle  a  roof  36  ft.  by  40  ft. 
if  shingles  are  laid  4^  in.  to  the  weather.     (Use  the  table  of  Art.  102.) 

Solution.     ~Tfj?r~  =  14.4  =  number  of  squares. 

833  XI 4. 4  =  11,995  =  number  of  shingles  required. 
/.  12,000  shingles  must  be  bought. 

6.  How  many  board  feet  in  26  pieces  of  2  in.  by  4  in.  by  14  ft.  long, 
20  pieces  of  3  in.  by  10  in.  by  16ft.  long?  Ans.  1043. 

6.  What  will  it  cost  at  $28  per  M  to  cover  the  floor  of  a  barn  32  ft. 
by  42  ft.  with  2-in.  plank?  Ans.  $75.26. 

7.  How  many  board  feet  are  there  in  3  sticks  of  timber  12  in.  by  14  in. 
and  22  ft.  long?  Ans.  924. 

8.  Find  the  total  cost  of  shingling  the  two  sides  of  a  roof  each  18  ft. 
by  40  ft.     Redwood  shingles  at  $4.75  a  thousand  are  used,  and  the  laying, 
nails,  etc.,  cost  $1.90  per  square.     Shingles  are  to  be  laid  5  in.  to  the 
weather.     (Use  the  table  of  Art.  102.)  Ans.  $79.61. 


124 


PR  A  CTICAL  MA  THEM  A  TICS 


9.  What  docs  the  following  cost  at  25  cents  a  foot: 

1  piece  |  in.  by    6  in.  by  10  ft. 
I  piece  }  in.  by    8  in.  by  12  ft. 

1  piece  I  in.  by  18  in.  by  4  ft. 

2  pieces  1  in.  by  6  in.  by  8  ft  ? 

Ant.  IG.75. 

10.  Find  the  cost  of  the  following  bill  of  lumber  if  the  quarter  sawed 
is  $90  per  M  and  the  common  sawed  is  $65  per  M : 

2  pieces  li  in.  X2J  in.  X12  ft.  quarter  sawed 
2  pieces  J  in.  X  8  in.  X 12  ft.  quarter  sawed 
1  piece  |  in.  X2J  in.  X12  ft.  quarter  sawed 
1  piece  f  in.  X  2  in.  X 12  ft.  quarter  sawed 

5  pieces    £  in.  X  3  in.  X 12  ft.  quarter  sawed 
1  piece      I  in.  XlO  in.  X 12  ft.  quarter  sawed 
1  piece      \  in.  X 10  in.  X  6  ft.  common  sawed 

6  pieces    i  in.  X  6  in.X  12  ft.  common  sawed 
4  pieces    i  in.  X  6  in.  X12  ft.  common  sawed. 

An*.  $0.18. 


Length  00; 
Roof  extending 
'2 'at  each  end 


Fio.  41. 

11.  Fig.  41  is  the  end  of  a  barn.     Find  the  area  of  one  end.     Find  the 
area  of  the  roof.     The  rafters  are  placed   16  in.  from  center  to  center. 
Find  the  number  of  board  feet  in  the  rafters  if  made  of  2  in.  X  6  in. 
(Use  12-ft.  stuff  for  short  rafters.)     Find  number  of  feet  of  lumber  to 
cover  ends,  sides  and  roof.     Find  how  many  shingles  it  will  take  for  the 
roof  if  laid  4J  in.  to  the  weather. 

Ans.  1310  ft.1;  3434J  ft.1;  2744;  8935;  28,750. 

12.  A  ship  builder  gave  $300  for  a  standing  oak  tree  to  make  a  long 
ship  timber.     The  cost  of  felling,  hewing,  and  hauling  was  $275.     If  the 
timber  was  18  in.  square  and  98  ft.  long,  find  the  number  of  board  feet 
in  it  and  the  cost  per  thousand  feet.  Ans.  2646;  $217.31. 


PLANE  SURFACES.     LINES  AND  ANGLES  125 

13.  Find  the  number  of  board  feet  in  the  following  list  of  framing  tim- 
ber for  a  house: 

Girders 5  pieces  6  in.  X  8  in.  X20  ft. 

Sills 16  pieces  6  in.  X  6  in.  X 16  ft. 

First  floor  beams 45  pieces  3  in.  XlO  in.  X28  ft. 

Second  floor  beams 45  pieces  3  in.  X  8  in.  X28  ft. 

Ribbons 16  pieces  1  in.  X  8  in.  X20  ft. 

Plates 32  pieces  2  in.  X  4  in.  X 16  ft. 

Outside  wall  studs 156  pieces  2  in.  X  4  in.  X20  ft. 

Inside  wall  studs 200  pieces  2  in.  X  4  in.  X 12  ft. 

Rafter  studs 90  pieces  2  in.  X  8  in.  X24  ft. 

Collar  beams 45  pieces  2  in.  X  6  in.  X 16  ft. 

Ans.   14,673. 


CHAPTER  XI 
TRIANGLES 

THE  RIGHT  TRIANGLE 

103.  A  right  triangle  is  a  triangle  having  one  right  angle. 
The  side  opposite  the  right  angle  is  called  the  hypotenuse, 
and  the  sides  about  the  right  angle  are  called  base  and  altitude, 
the  base  being  the  side  the  triangle  is  supposed  to  rest  upon. 

The  right  triangle  is  of  great  importance  as  it  is  of  very  com- 
mon occurrence  in  practice.  The  solution  of  the  right  triangle 

depends  upon  the  following  relation  es- 

tablished in  geometry. 

The  square  formed  on  the  hypotenuse  is 

equal  to  the  sum  of  the  squares  formed  on 

the  other  two  sides. 

This  may  be  illustrated  as  in  Fig.  42. 

AC  is  the  hypotenuse  and  is  5  units  in 

length.     AB  is  the  base,  4  units  long. 

BC  is  the  altitude,  3  units  long.     Here  it 

is  easily  seen  that  the  square  on  AC  is 
equal  to  the  sum  of  the  squares  on  AB  and  BC.  Hence 
AC2  =  AB2+BC2,  or  in  general,  if  c  stands  for  the  hypotenuse, 
6  for  base,  and  a  for  altitude,  then  c2  =  a2+62.  From  this 
are  derived  the  three  following  formulas,  by  which  any  side 
can  be  found  if  the  other  two  are  known. 


Fio.  42. 


Example.     Find  the  hypotenuse  of  a  right  triangle  whose 
base  is  14  ft.  and  altitude  16  ft. 

Solution.     Using  formula  [9],     c  =  \/a2+&2. 


=  Vl62+142=\/452  =  21.26+  ft.  Ana. 
126 


TRIANGLES 


127 


EXERCISES  33 

In  the  following  right  triangles,  solve  for  the  parts  named  in  the  exer- 
cise: 

1.  a  =  25,  6  =  16,  find  c  and  area. 

Ans.  c  =  29.68;  area  =200  square  units. 

2.  c  =  46,  6=30,  find  a  and  area. 

Ans.  a  =  34. 87  +  ;  area  =  523. 05+  square  units. 

3.  Area  =2  acres,  a  =  15  rd. ;  find  b  and  c. 

Ans.  b  =  42.667-  rd. ;  c=  45.23-  rd. 

4.  a  =  16,  c  =  20,  find  b  and  area. 

Ans.  b  — 12;  area  =96  square  units. 

5.  Find  length  of  the  diagonal  of  a  rectangle  16  ft.  by  14  ft. 

Ans.  21.26ft. 


6.  Find  the  diagonal  of  a  cube  9  ft.  on  an  edge. 
Suggestion.     In  Fig.  44,  the  line   marked   D  is 
of  the  cube.     First  find  d  and  then  D. 


Ans.  15.588+  ft. 
called   the   diagonal 


d=\/92+92=Vl62. 

D=\/162+92=\/243. 

=  A/8lX3=9\/3. 

7.  A  man  swims  at  right  angles  to  the  bank  of  a  stream  at  the  rate  of 
3.5  miles  per  hour.     If  the  current  is  7.5  miles  per  hour,  find  the  rate 
the  man  is  moving.  Ans.  8.28—  miles  per  hour. 

Suggestion.     The  rate  the  man  is  moving  is  the  length  of  the  hypote- 
nuse of  a  right  triangle  having  a  base  =3. 5  mi.  and  an  altitude  =  7. 5  mi. 

8.  The  diagonal  of  a  rectangle  is  130  and  the  altitude  is  32.     Find  the 
area.  Ans.  4032  square  units. 

9.  What  is  the  length  of  the  longest  line  that  can  be  drawn  within  a 
rectangular  box  12  ft.  by  4  ft.  by  3  ft.?  Ans.  13  ft. 

10.  The  hypotenuse  of  a  right  triangle,  with  base  and  altitude  equal, 
is  12  ft.     Find  the  length  of  the  base  and  altitude.         Ans.  8.485+  ft. 

11.  The  base  of  a  triangle  is  20  ft.  and  the  altitude  is  18  ft.     What  is 
the  side  of  a  square  having  the  same  area?  Ans.  13.416+  ft. 

12.  The  area  of  a  rectangular  lawn  is  5525  m.8,  and  the  length  of  one 
of  its  sides  is  8.5  Dm.     Find  the  length  of  its  diagonal  in  meters  to  three 
decimal  places.  Ans.  107.005—  m. 


128 


PRACTICAL  MA  Til  KM  A  TICK 


18.  A  steamer  goes  due  north  at  the  rate  of  15  miles  per  hour,  and 
another  due  west  at  18  miles  per  hour.  If  both  start  from  the  same  place, 
how  far  apart  will  they  be  in  6  hours?  Ana.  140.58  +  miles. 

14.  What  is  the  length  of  the  diagonal  of  a  room  20  ft.  by  16  ft.  by 
12  ft.?  Ann.  28.284+  ft. 

15.  Find  cost  at  $20  per  M  of  roof  boards  on  a  third-pitch  roof  of  a 
barn  45  ft.  by  65  ft.,  if  projections  at  ends  and  eaves  are  2  ft.     (In  a 
third-pitch  roof  the  distance  of  the  ridge  above  the  plate  is  one-third 
the  width  of  the  building.)  Ans.  $80.15. 

Suggestion.     Distance  of  ridge  above  plate  =  J  of  45  ft.  =  15  ft. 
Length  of  rafters  without  projection  =  V22.5* +  15*  =  27.04  ft. 
Total  length  of  rafters  =2  ft.  +27.04  ft.  =29.04  ft. 
Area  of  one  side  of  roof  =  29.04X69  =  2003.76  sq.  ft. 

16.  How  many  thousand  shingles  will  it  take  to  cover  the  above  roof, 
if  shingles  are  laid  4J  in.  to  the  weather  and  a  double  row  is  put  at  the 
beginning  on  each  side?     (No  allowance  for  waste.) 

Ans.  32,500  nearly. 

17.  In  fitting  a  steam  pipe  to  the  form  ABCD,  Fig.  45,  making  a  bend 
of  45°,  the  fitter  takes  BC  =  CE  +  &CE.     What  is  the  error  if  CE  =•  18  in.? 
What  is  the  correct  length  of  CB,  and  what  is  the  per  cent  of  error  by 
the  fitter's  method? 

Ans.  Error,  0.0442- ;  correct,  C#  =  25.4558  +  in.;  %  of  error,  0.17  +  . 


18.  In  cutting  a  rafter  for  a  half-pitch  roof  a  carpenter  makes  the 
length  of  the  rafter  AB  =  1  ft.  5  in.  for  every  foot  there  is  in  AC,  Fig.  46. 
If  AC  =8  ft.,  find  A B  by  this  rule.     What  is  the  per  cent  of  error  by  this 
method?        Ans.  Carpenter's  method,    11  ft.   4  in.;  correct,  11  ft. 

3.76+  in.;  error,  0.17+  %. 

19.  To  find  the  diagonal  of  a  square,  multiply  the  side  by  10,  take  away 
1  %  of  this  product,  and  divide  the  remainder  by  7.     Test  the  accuracy 
of  this  rule.  Ans.  0.006  —  %  too  large. 

Solution.     Take  a  square  with  a  side  of,  say,  25  in. 

10X25   =250 
1%  of  250  =     2.5 
Remainder  =  247.5 

247.5 -J- 7  =  35.357  +  in.  -  diagonally  jrule. 
By  formula  for  hypotenuse,  diagonal  =  v/25*+ 25*  =  35.355  +  in. 
Hence  error  -35.357  in.  -35.355  in.  -0.002  in. 

0.002  4-  35.355  =  0.006-  %  =per  cent  of  error. 


TRIANGLES 


129 


It  is  evident  that  this  rule  is  very  accurate  and  is  also  easy  of  applica- 
tion. 

20.  Show  that  the  following  rules  are  correct.     They  are  very  useful 
in  many  problems  connected  with  a  square. 

(1)  The  diagonal  of  a  square  equals  a  side      T 
of  the  square  multiplied  by  \/2. 

(2)  The  side  of  a  square  equals  one-half  the 
diagonal  multiplied  by  \/2. 

The  number  of  decimal  places  used  in  \/2 
will  depend  upon  the  degree  of  accuracy 
desired.  Pipe  fitters  usually  use  \/2  =  1.41. 
It  is  often  necessary  to  take  three  or  more 
decimal  places.  \/2  =  1.4142136  to  seven 
decimal  places. 

21.  Use  rule  (1)  in  obtaining  the  correct  values  in  exercises  17,   18, 
and  19. 

22.  What  is  the  distance  across  the  corners  of  a  square  nut  that  is 
3s  in.  on  a  side?     Use  rule  (1). 


FIG.  48. 


FIG.  49. — Cap  screw. 


23.  What  must  be  the  diameter  of  round  stock  so  that  a  square  bolt 
head  If  in.  on  a  side  may  be  milled  from  it? 

24.  Find  the  distance  across  the  flats  of  the  square  head  of  a  cap  screw 
that  may  be  milled  from  round  stock  1|  in.  in  diameter.     Use  rule  (2) 
of  exercise  20. 


Plan 


Section 


FIG.   50. 


26.  Fig.  50  shows  a  "scissors"  roof  truss  with  the  lengths  AB  =  B(  = 
AC  =  30  ft.,  CD  =  CF  =  1S  ft,,  and  CG  =  CE=8  ft,  Find  the  lengths  of 
NC  and  FG.  Ans.  NC  =  25  ft.  11  f  in. ;  FG  =  13  ft.  10£  in. 


130 


J'KA  CTICAL  MA  THEM  A  TICS 


26.  A  smokestack  is  held  in  position  by  three  guy  wires  that  reach 
the  ground  49  ft.  from  the  foot  of  the  stack.  Find  the  length  of  a  guy 
wire  if  they  are  fastened  to  the  stack  70  ft.  from  the  ground. 

Ans.  85.4+  ft. 

27.  An  engine  shaft  is  centered  9  ft.  below 
and  3  ft.  to  the  left  of  the  center  of  a  line 
shaft.      Find  the  distance  between  the  cen- 
ters of  the  two  shafts.       Ans.  9  ft.  5  j  in. 

28.  The    dimensional    sketch,    Fig.    51, 
shows  plan  and  section  of  a  roof.     It  has  to 
be  boarded.      What  will  be  the  number  of 
feet  of  boards  required?  Ans.  356. 

29.  In  the  gambrel  roof  shown  in  section 
in  Fig.   52,   find  the   lengths  of  rafters  and 
parts  not  given. 

DB  =  15  ft.  21  in.;  £C  =  14  ft.  3f  in.     All  to 


Fio.  52. 


Ans.  AB  =  15  ft.  71  in.;  DB 
the  nearest  |  in. 

SIMILAR  TRIANGLES 

104.  Triangles  that  have  the  same  shape  are  said  to  be 
similar. 

In  Fig.  53 (a),  ABC  and  ADE  are  similar.     In  (b)  ABC  and  A'B'C' 

arc  similar. 


Draw  two  triangles  as  in  (a)  and  measure  the  sides  a,  b,  c 
and  a',  b',  c'.  Then  determine  the  ratios  a  :b,  arc,  b  :  c, 
a1  :  b',  a'  :  c'  and  6'  :  c'.  Follow  the  same  directions  for 
the  triangles  in  (6).  Now  compare  the  values  of  the  ratios 
and  notice  whether  or  not  they  are  equal. 

The  results  of  the  above  should  lead  to  the  following: 
a  :  b  =  a'  :  b';  a  :  c  =  a'  :  c';  and  b  :  c  =  b'  :  c'. 


TRIANGLES 


131 


Two  triangles  that  have  the  angles  of  one  equal  respectively 
to  the  angles  of  the  other  are  smilar. 

The  sides  about  the  equal  angles  in  the  similar  triangles 
are  called  corresponding  sides. 

Thus,  c  and  c'  are  corresponding  sides.  Other  corresponding  sides  are 
a  and  a',  and  b  and  &'. 

From  the  proportions  given  above,  we  arrive  at  the  fol- 
lowing principle: 

Corresponding  sides  of  similar  triangles  form  a  proportion. 

Example.  To  find  the  distance  between  the  points  P  and  Q 
on  opposite  banks  of  a  stream,  Fig.  54,  where  P  is  inaccessible. 


FIG.   54. 

Solution.  As  shown  in  the  figure,  measure  distances  AQ  = 
16  ft.,  AB  =  10  ft.,  and  BC  =  60  ft.  Because  triangles  AQB 
and  APC  are  made  similar  we  have  the  proportion 

AB  :AQ  =  AC  :  AP 

.MO  ft.  :  16  ft.  =  70  ft.  :AP 


.„ 
.  .AP  = 


16X70 


I10,, 
=  112  ft. 


.'.PQ=  112  ft.-  16  ft.  =96  ft.    Ans. 

105.  Tapers.  —  The  man  in  the  machine  shop  often  finds 
it  necessary  to  determine  the  taper  per  foot  of  a  piece  that 
is  to  be  turned,  in  order  that  he  may  set  his  lathe  properly. 
By  the  taper  per  foot  is  meant  the  decrease  in  diameter 
if  the  piece  is  1  ft.  long. 

In  Fig.  55  (a),  the  taper  is  evidently  4£  in.  —  3  in.  =  1^  in, 
per  foot.  In  (6)  the  taper  is  2\  in.  —  2  in.  —  \  in.  in  4  in. 
Hence  the  taper  per  foot  is  3  times  as  much  or  \\  in. 


132  PRACTICAL  MATHEMATICS 

If  I  stands  for  the  length  of  tapered  part  in  feet,  t  for  the 
taper  in  inches  in  this  part,  and  T  for  the  taper  in  inches  per 
foot,  then  the  following  proportion  is  true  by  similar  triangles: 


The  taper  for  the  total  length  of  the  piece  is  evidently 
the  taper  per  foot  times  the  length  in  feet. 

Example.     In    Fig.    55  (c),    what   is   the   taper   per   foot? 
What  would  be  the  taper  for  total  length  of  piece? 

Solution.     Substituting  in  [12],  T\  :  1  =\  :  T. 

.•.r=lXi-5-T87=lTin.     Ans. 
1J  in.Xvf  =  2.3  in.  =  taper  if  it  were  tapered  the  full  length. 


r< 10 


K  —  8  ---  >j<-4->i 

Fro.  55. 

106.  Turning.  —  In  turning  a  piece  in  a  lathe  the  taper  is 
sometimes  made  by  shifting  the  tailstock  of  the  lathe.  Since, 
when  the  piece  is  revolved,  the  same  cut  is  made  on  all  sides,  it 
is  necessary  to  set  the  tailstock  over  one-half  of  what  the  taper 
would  be  if  the  piece  were  tapered  the  full  length.  Thus,  a 
piece  1  ft.  long  with  a  taper  lj  in.  per  foot  requires  the  tail- 
stock  to  be  set  over  \  of  \\  in.  =  |  in. 

If  Z)  =  the  large  diameter  and  d  the  small  diameter  of 
the  tapered  portion,  L  the  total  length  of  the  piece,  and  / 
the  length  of  the  tapered  portion,  then  the  offset  x  of  the  tail- 
stock  is  determined  by  the  following  formula: 


Example.  A  shaft  3  ft.  long  is  to  have  a  taper  turned  on 
one  end  10  in.  long,  the  large  end  of  the  taper  being  4  in. 
in  diameter  and  the  small  end  3$  in.  Find  the  distance  to 
set  over  the  tailstock. 

Solution.     x  =         -*X      =  0.9in.     Ans. 


TRIANGLES  133 

EXERCISES  34 

1.  The  following  tapers  per  inch  are  what  tapers  per  foot:  0.0026  in.; 
0.0473  in.;  0.0379  in.? 

2.  How  much  will  the  tailstock  need  to  be  set  over  to  give  a  taper  of 
1|  in.  per  foot  if  the  work  is  1  ft.  in  length?     If  8  in.  in  length? 

Am.   &  in.;  f  in. 

3.  The  standard  pipe  thread  taper  is  f  in.  per  foot.     How  much  is 
this  per  inch?  Ans.  -^  in. 

4.  Find  the  taper  per  foot  to  be  used  in  turning  a  pulley  with  a  14  in. 
face  crowned  j%  in.  Ans.  0.32+  in. 

5.  If  the  crowning  of  a  pulley  is  ->\  of  the  width  of  the  face,  find  the 
taper  per  foot  to  be  used  in  turning  a  pulley  with  a  10-in.  face. 

Ans.   I  in. 


Taper  pin.     -Taper  J  in.  per  foot. 


Taper-pin  reamer.     Taper  \  in.  per  foot. 
FIG.   56. 

6.  A  taper-pin  reamer  has  a  taper  of  J-  in.  per  foot.     If  the  diameter 
of  the  small  end  is  0.398  in.  and  the  length  of  the  flutes  is  5j  in.,  find  the 
diameter  of  the  large  end  of  the  flutes. 

7.  A  taper  reamer  has  a  taper  of  f  in.  per  foot  and  the  flutes  are  3£  in. 
long.     If  it  is  |  in.  in  diameter  at  the  large  end,  find  the  diameter  at 
the  small  end.  Ans.  0.5677  in. 

8.  Find  the  taper  per  foot  of  a  taper  reamer  that  has  a  diameter  of 
f?  in.  at  the  large  end  of  the  flutes,  and  ff  in.  at  the  small  end  if  the  flutes 
are  2f  in.  long.  Ans.  f  in. 

9.  A  taper  reamer  has  a  taper  of  f  in.  per  foot.     If  the  diameter  of  the 
large  end  is  1^,  find  the  diameter  of  the  small  end,  the  flutes  being  3f  in. 
long.  Ans.  fi  in. 

10.  In  Fig.  57,  the  timbers  CB,  DF,  and  EG  are  perpendicular  to  CA. 
From  the  given  dimensions  find  the  lengths  of  DF,  EG,  CF,  DG,  and 
AB.  Ans.  EG  =  4  ft.;  DG  =  10.770  ft.;  AB  =  32.311  ft. 

Suggestion.     CA  :CB  =  DA:  DF. 
Or  30:  12  =  20  :  D/*1. 


30 
CF  =  v/io'^XS1  =  Vl64  =  12.800 


134 


PRACTICAL  MA  THEM  A  TICS 


11.  How  much  should  the  tailstock  be  offset  to  turn  a  taper  on  a 
piece  of  work  10  in.  long,  if  the  tapered  portion  is  4  J  in.  long  and  measures 
1.275  in.  in  diameter  at  the  large  end  and  0.845  in.  at  the  small  end? 

Ana.  0.5212  in. 

12.  What  is  the  offset  of  the  tail  center  for  turning  a  taper  18  in.  long 
on  a  bar  26  in.  long,  if  the  diameters  at  the  ends  of  the  taper  are  3}  in. 
and  2}  in.?  Ana.  0.27  +in. 


10' 


D         WE 
Fiu.  57. 


13.  To  find  the  distance  across  the  lake  in  Fig.  58,  measure  AB  =  80  rd.  ; 
AD  =30   rd.  ;    DE  =  25  rd.,   and    find    BC.     Is   it   necessary   to   make 
right-angled  triangles?  Ans.  66§  rd.;  no. 

14.  Show  how  to  find  the  height  of  a  smokestack  CD  of  Fig.  59,  when 
the  foot  of  the  stack  cannot  be  reached. 


Suggestion.  On  a  level  place  measure  from  A  to  B  in  a  line  with  C, 
and  measure  angles  BAD  and  CBD.  Suppose  the  line  AB  is  40  ft.  and 
the  angles  are  40°  and  60°  respectively.  Construct  a  figure  to  scale  on 
paper  and  measure  the  line  that  corresponds  to  the  smokestack. 


THE  STEEL  SQUARE 

107.  One  of  the  most  useful  instruments  known  to  man  is 
the  ordinary  steel  square  or  carpenter's  square  shown  in 
Fig.  60.  It  is  made  in  various  sizes  but  the  most  common 
size  is  with  the  longer  arm,  called  the  body,  blade,  or  stock, 


TRIANGLES  135 

24  in.  in  length  and  2  in.  in  breadth,  and  the  shorter  arm, 
called  the  tongue,  16  in.  or  18  in.  in  length  and  1£  in.  in 
breadth. 

Many  books,  having  in  some  cases  five  or  six  hundred  pages, 
have  been  written  on  the  uses  of  the  steel  square.  Here  we 
wish  only  to  call  attention  to  the  fact  that  the  principles 
involved  in  using  the  steel  square  are  mainly  those  involved 
in  the  solution  of  the  right  triangle  and  in  similar  triangles. 
One  who  understands  the  right  triangle  can  devise  many  uses 


FIG.  60.  FIG.  61. 

for  the  steel  square,  and  can  readily  see  the  principles  under- 
lying the  various  uses  of  this  instrument  given  in  the  treatises 
on  the  steel  square. 

Upon  the  ordinary  steel  square  are  found  many  figures, 
telling  lengths  of  braces,  board  measures,  etc.  No  attempt 
will  be  made  here  to  explain  these. 

Example.  By  use  of  a  steel  square,  find  the  length  of  the 
hypotenuse  of  a  right  triangle  that  has  a  base  of  8  in.  and  an 
altitude  of  7  in. 

Solution.  Measure  the  line  drawn  from  the  8-in.  mark 
on  the  blade  to  the  7-in.  mark  on  the  tongue.  This  measures 
about  lOf  in.  which  is  near  enough  for  most  practical  purposes. 
By  the  right  triangle  method  the  hypotenuse  =  -\/8~2+V2  = 
10.63 +in.  (See  Fig.  61.) 

This  method  can  readily  be  applied  to  find  the  lengths  of 
braces  supporting  two  pieces  that  are  perpendicular  to  each 
other;  to  find  rafter  lengths,  lengths  of  the  parts  of  a  trestle, 
etc. 

108.  Rafters  and  roofs. — The  run  of  a  rafter  is  the  distance 
measured  on  the  horizontal  from  its  lower  end  to  a  point  under 


130 


PRACTICAL  MATHEMATICS 


its  upper  end.     The  rise  is  the  distance  of  the  upper  end  above 

the  lower  end.     In  Fig.  62,  AC  is  the  run  and  CB  the  rise. 
The  slant  of  a  roof  is  usually  told  by  stating  the  relation  of  the 

rise  to  the  run.     It  is  often  given  by  stating  the  rise  per  foot 

of  run;  as  6  in.  to  1  ft. 
Another  way  is  to  state 
what  is  known  as  the 
pitch  of  the  roof.  A  roof 
is  said  to  be  half  pitch, 
D  quarter  pitch,  full  pitch, 
etc.,  when  the  rise  is  \, 
},  1,  etc.,  times  the  full 
width  of  the  building  as 

represented  in  Fig.  62,  where  AD  is  the  width  of  the  building. 
The  relation  between  rise  and  pitch  is  shown  by  the  following 

table: 


4  ft.  rise  is  £  pitch. 
6  ft.  rise  is  \  pitch. 
8  ft.  rise  is  \  pitch. 
10  ft.  rise  is  VV  pitch. 
12  ft.  rise  is  \  pitch. 
15  ft.  rise  is  f  pitch. 
18  ft.  rise  is  f  pitch. 


12  ft.  run  to 
12  ft.  run  to 
12  ft.  run  to 
12  ft.  run  to 
12  ft.  run  to 
12  ft.  run  to 
12  ft.  run  to 
12  ft.  run  to  24  ft.  rise  is  full  pitch. 


109.  Uses  of  the  square. 
-The  bevel  or  slant  on 
the  end  of  a  brace  or 
rafter,  necessary  to  make 
it  fit  the  part  it  rests 
against,  can  easily  be 
marked  by  the  square.  LL 

Example  \.  Required  to 
cut  the  lower  end  of  a 
rafter  that  is  to  rest  on  the 
plate  if  the  rise  of  the  rafter  is  8  ft.  and  the  run  12  ft. 

Discussion.  Place  the  square  as  shown  in  Fig.  63  and  mark 
along  the  lower  edge.  This  gives  the  proper  slant.  Marking 
along  the  tongue  gives  the  slant  for  the  upper  end  of  the  rafter. 


FKI.  03. 


TRIANdLEti 


137 


In  placing  the  square  on  the  stick  it  is  only  necessary  to  take 
the  distances  on  the  blade  and  tongue  in  the  same  ratio  as  the 
ratio  of  the  run  to  the  rise.  In  this  case  we  could  as  well  have 
taken  24  in.  and  16  in.  or  9  in.  and  6  in. 

Example  2.  Required  to  cut  a  rafter  for  a  V-shaped  roof 
on  a  building  12  ft.  wide  if  the  rise  of  the  rafter  is  to  be  4  ft. 
The  rafter  is  to  be  made  of  a  piece  of  2  in.  by  4  in.  and  half  its 
width  is  to  project  18  in.  beyond  the  plate. 


FIG.  04. 

Discussion.  Determine  the  slant  for  the  plate  end  as  de- 
scribed in  example  1.  Then  place  the  square  as  shown  in 
Fig.  64  so  as  to  give  a  run  of  24  in.  to  a  rise  of  16  in.  The 
square  is  replaced  with  point  A  on  C  and  this  repeated  as 
often  as  necessary  to  give  a  run  of  half  the  width  of  the  building. 
In  this  case,  it  is  necessary  to  place  the  square  three  times.  In 
the  last  position,  a  mark  along  the  tongue  gives  the  slant  of  the 
upper  end  of  the  rafter  and  determines  the  length  of  the  rafter. 
Any  rafter  can  be  cut  in  this  way. 

EXERCISES  35 

1.  Show  with  a  carpenter's  square  how  to  determine  the  length  of  a 
brace  for  a  run  of  4  ft.  6  in.,  and  a  rise  of  3  ft.  6  in.     Show  how  to  cut 
bevels  on  ends. 

2.  Show  how  to  mark  the  slants  for  the  legs  of  the  sawhorse  shown  in 
Fig.  65.     Compute  the  length  of  legs.  Ans.  29|  in.  nearly. 

3.  In  Fig.  66  is  a  plan  of  one  end  of  a  roof  on  a  house  18  ft.  in  width 
and  28  ft.  long.     CB,  DE,  etc.,  are  common  rafters;  AB  and  NB  are  hip 
rafters;  and  FO,  HI,  etc.,  are  jack  rafters.     Find  the  lengths  to  cut  the 
several  rafters  if  the  roof  is  £  pitch.     Show  how  to  determine  the  slants 


138 


PRACTICAL  MATHEMATICS 


at  both  ends  of  each.     The  rafters  do  not  extend  beyond  the  plates  and 
are  placed  1  ft.  6  in.  from  center  to  center. 

4.  Find  lengths  of  the  common,  hip,  and  jack  rafters  for  the  roof  of 
which  Fig.  67  is  the  plan.  It  is  J  pitch,  rafters  1  ft.  6  in.  from  center  to 
center,  and  extend  2  ft.  beyond  the  plates. 


Fio.  69. 

6.  Fig.  68  is  a  plan  of  the  roof  of  a  hexagonal  tower.  Find  lengths  of 
the  rafters  that  end  at  the  plates.  Full  pitch  and  rafters  1  ft.  6  in. 
between  centers.  Width  of  tower  is  12  ft. 


TRIANGLES 


139 


6.  Fig.  69  represents  the  plan  of  a  roof  of  a  house  20  ft.  square,  with 
a  flat  circular  portion  8  ft.  in  diameter.  If  the  circle  is  one-third  the 
width  of  the  building  above  the  plates  and  the  rafters  2  ft.  between  centers 
on  the  plates,  find  the  length  of  each  rafter  and  show  how  to  cut  slants. 


ISOSCELES  AND  EQUILATERAL  TRIANGLES 

110.  Two  other  forms  of  triangles  of  common  occurrence 
are  isosceles  and  equilateral  triangles. 

A  triangle  which  has  two  equal  sides  is  called  an  isosceles 
triangle. 

A  triangle  which  has  all  of  its  sides  equal  is  called  an  equi- 
lateral triangle. 

The  following  facts  are  proved  in  geometry.  The  student 
should  satisfy  himself  that  they  are  true  by  constructing  the 
figures  and  measuring  the  parts. 

111.  The    isosceles    triangle. — In    Fig.    70    the    isosceles 
triangle  ABC  has  equal  sides  AC  and  BC. 

The  angles  A  and  B  opposite  the  equal  sides  are  equal. 


The  line  CD  drawn  bisecting  the  vertex  angle,  C,  is  per- 
pendicular to  and  bisects  the  base,  AB.  That  is,  AD  =  DB. 
It  also  divides  the  isosceles  triangle  into  two  equal  right  trian- 
gles, BDC  and  ADC. 

The  line  CD  is  then  the  bisector  of  the  angle  C,  and  also  a 
median  and  an  altitude  of  the  triangle. 

The  diagonal  of  a  square  divides  the  square  into  two  equal 
right  isosceles  triangles.  In  these  isosceles  triangles  each  of  the 
equal  angles  is  45°. 

112.  The  equilateral  triangle.— In  Fig.  71,  triangle  ABC  has 
its  three  sides  equal  and  is  an  equilateral  triangle. 


140 


PRACTICAL  MA  THEM  A  TICS 


The  angles  opposite  the  equal  sides  are  equal,  and  therefore 
each  angle  equals  60°. 

The  line  drawn  from  the  vertex  A  and  bisecting  the  angle  is 
perpendicular  to  and  bisects  the  opposite  side  BC.  It  also 
divides  the  equilateral  triangle  into  two  equal  right  triangles, 
ABD  and  ADC. 

Furthermore,  each  of  the  lines  BE  and  CF  divides  the  trian- 
gle in  the  same  manner  that  it  is  divided  by  AD. 

Each  of  these  lines  then  is  a  bisector  of  an  angle,  and  also  a 
median  and  an  altitude  of  the  equilateral  triangle. 


E 


D 


FIG.  71. 


The  point  0  where  these  three  lines  meet  is  called  the 
center  of  the  equilateral  triangle.  It  is  one-third  the  distance 
from  one  side  to  the  opposite  vertex.  That  is,  DO  =  \DA, 


It  follows  then  that  AO  =  2DO,  BO  =  2EO,  and  CO  =  2FO. 

Either  of  the  triangles  formed  when  an  equilateral  triangle 
is  divided  into  two  triangles  by  an  altitude  is  a  right  triangle 
having  acute  angles  of  30°  and  60°.  This  triangle  is  very 
important  in  practical  work.  It  is  readily  seen  that,  in  such 
a  right  triangle,  the  hypotenuse  is  twice  the  shortest  side. 
That  is,  NR  =  2MR. 

An  altitude  of  an  equilateral  triangle  equal*  one-half  of  a 
side  times  \/3.  As  a  formula,  a  =  $s\/3  =  i.sX1.732,  where 
a  =  an  altitude  and  s  =  a  side. 

This  is  obtained  by  solving  the  right  triangle  AFC  for  FC. 
The  student  can  easily  carry  through  the  work  for  any  par- 
ticular value  of  a  side.  The  following  is  the  derivation  in 
general  and  involves  some  algebra. 

Let  s  stand  for  one  side  of  the  equilateral  triangle. 

Then   /•'(  '  =  VJC*-JF*  =  vV-($6>)2 


TRIANGLES  141 

From  this  it  follows  that: 

A  side  of  an  equilateral  triangle  equals  twice  the  altitude 
divided  by  \/3.  As  a  formula,  s  =  2a-r-  \/3- 

Since  the  area  of  any  triangle  is  one-half  its  base  times  its 
altitude,  it  follows  that : 

The  area  of  an  equilateral  triangle  equals  the  square  of  one-half 
a  side  times  \/3.  As  a  formula,  A  =  (|s)2-v/3  =  (^s)2X  1.732, 
where  A  is  the  area. 

113.  The  regular  hexagon. — Another  form  often  used  in 
practice  is  the  regular  hexagon.  From  geometry  we  learn 
the  following  facts  which  will  ap- 
pear true  from  a  careful  considera- 
tion of  Fig.  72.  The  diagonals, 
drawn  as  shown,  divide  the  hexa- 
gon into  six  equal  equilateral 
triangles.  The  distance  from  the 
center  0  to  any  vertex  is  the  same 
as  the  length  of  a  side.  The  area 
of  the  regular  hexagon  is  equal  to 
six  times  the  area  of  an  equilateral 
triangle  with  sides  equal  to  the 

sides  of  the  hexagon.  The  altitude  NO  may  be  found  by 
solving  the  right  triangle  ANO;  or  may  be  found  by  taking 
AN  times  \/3. 

EXERCISES  36 

1.  Find  the  altitude  of  an  isosceles  triangle  whose  equal  sides  are 
16  ft.  and  base  14  ft.  Ans.   14.387+  ft. 

2.  Find  the  base  of  an  isosceles  triangle  if  equal  sides  are  18  in.  and 
altitude  16.5  in.  Ans.   14.387+  in. 

3.  Find  the  altitude  of  an  equilateral  triangle  the  sides  of  which  are 
12ft.  Ans.  10.392+  ft. 

4.  In  Fig.  72,  find  ON  if  AB  is  10  ft.  Ans.  8.660+  ft. 

5.  Find  the  area  of  an  equilateral  triangle  18  in.  on  a  side. 

Ans.  140.3-  in.2 

6.  Find  the  area  of  an  isosceles  triangle  whose  base  is  6  vn.  and  equal 
sides  9  in.  .  Ans.  25.456— in.2 

7.  Compute  the  area  of  a  regular  hexagon  one  of  whose  sides  is  5  ft. 

Am.  64.95+  ft.2 

8.  An  equilateral  triangle  has  an  area  of  21.217  in.2     What  is  the 
length  of  one  side?  Ans.  7  in. 


\\'2 


PRACTICAL  MATHEMATICS 


Solution.     Ou  page  141  it  is  stated  that  the  area  of  an  equilateral 
triangle  equals  the  square  of  one-half  a  side  times  \  :•;. 
Or  A  -da)1  XI.  732. 

Then  J« 


•*•!  .732. 
But  A  is  given  equal  to  21.217. 

•'•  i«  -\/21.217  -5-1.732  -Vl2i25  -3.5. 
Or  8  =  2X3.5  =  7. 

9.  An  isosceles  triangle  has  a  base  16  in.  long  and  the  equal  sides  18 
in.     What  is  the  area?  Ans.  128.996+  in.' 

10.  Find  the  length  of  the  steam  pipe  ABCD,  Fig.  73,  if  AD  =  6  ft., 
C£  =  16  in.,  and  angle  EEC  =  30°.  Ans.  76.29-  in. 


B 


E 
FIG.  73. 


Suggestion.     ^  BC  =  2EC  =  2  X 16  in.  =  32  in. 

"  BE  =  \/3XEC  =  1.732X16  =  27.71+  in. 

Increase  along  #C  =  32  in. -27.71+  in.  =4.29-  in. 

Length  of  ABCD  =  72  in. +4.29-  in.  =76.29-  in. 

11.  The  hypotenuse  of  a  right  triangle  is  5  ft.  and  one  side  is  4  ft. 
Show  that  the  equilateral  triangle  made  on  the  hypotenuse  is  equal  to 
the  sum  of  the  equilateral  triangles  made  on  the  other  two  sides. 

Suggestion.  Find  the  areas  of  the  three  triangles  and  show  that  the 
sum  of  the  areas  of  the  two  smaller  triangles  equals  the  largest  triangle. 
Use  the  formula  A  =  (J«)2X  1.732. 


Isosceles 
Right  Triangle 


80-60   Right 
Triangle 


Fio.  74. 

12.  Two  very  convenient  forms  of  triangles  used  by  draftsmen  are 
right  triangles  made  of  celluloid  or  rubber;  one  a  right  isosceles  triangle 
having  the  acute  angles  each  45°,  and  one  having  acute  angles  of  30°  and 
60°.     If  one  of  the  equal  sides  of  the  isosceles  right  triangle  is  6  in.,  find 
the  hypotenuse.  Ans.  8.485  in. 

13.  If  the  shortest  side  of  the  30°-60°-right  triangle  is  4  in.  find  the 
other  sides.  Ans.  6.928  in.  and  8  in. 


TRIANGLES 


143 


14.  Using  the  draftsman's  triangles,  show  how  to  construct  the  follow- 
ing angles:  15°.  75°,  105°,  120°,  135°,  150°. 

15.  A  hexagonal  nut  for  a  Q-in.  screw  is  lj  in.  across  the  flats.     Find 
the  diagonal,  or  the  distance  across  the  corners,  of  such  a  nut. 

Ans.  1.443  in. 


16.  What  is  the  distance  across  the  corners  of  a  hexagonal  nut  thai  is 
f  in.  on  a  side?     What  is  the  distance  across  the  flats  of  the  same  nut? 

Ans.   1.5  in.;  1.299  in. 

17.  Show  that  the  distance  across  the  corners  of  a  hexagonal  nut  is 
approximately  1.15  times  the  distance  across  the  flats. 

Suggestion.  It  is  readily  seen  that  the  distance  across  the  corners  is 
twice  the  side  of  an  equilateral  triangle  whose  altitude  is  one-half  the 
distance  across  the  flats. 

By  Art.  112,    s=2a +V1[. 

/.  2s  =  4a^v^3  =2aXl.l5. 

But  2s  =  distance  across  corners, 

and  2a  =  distance  across  flats. 

18.  In  a  standard  hexagonal  bolt  nut  the  distance  across  the  flats  is 
given  by  the  formula  F  =  1.5Z)  +  |,  where  F  is  the  distance  across  the 
flats  and  D  the  diameter  of  the  body  of  the  bolt. 

Find  the  distance  across  the  flats  and  across  the  corners  on  a  hexagonal 
nut  for  a  bolt  If  in.  in  diameter.  Ans.  2.75  in.;  3.1754  in. 

19.  To  what  diameter  should  a  piece  of  stock  be  turned  so  that  it 
may  be  milled  to  a  hexagon  and  be  If  in.  across  the  flats? 

Ans.  2.0207  in. 

20.  Find  the  size  of  round  stock  to  make  bolts  with  hexagonal  heads 
and  having  bodies  of  the  following  diameters:  f  in.,  1J  in.,  li  in.,  1^  in. 

Ans.  1.0104  in.;  2.7424  in.;  2.0929  in.;  2.4177  in. 
Suggestion.     Distance  ocross  flats,  ^  =  1.50+1. 
When  D  =  |  in.,  F  =  1.5X0.5+0.125  =  0.875  in. 

0.875X2 
Diameter  of  stock  =  distance  across  corners  =       ~7^=      =1.0104  in. 

By  the  rule  of  exercise  17,  distance  across  corners  =  0.875 XL  15  = 
1.006  in. 

SCREW  THREADS 

114.  In  the  United  States  there  are  in  use  several  different 
kinds  of  screw  threads.  Here  we  will  consider:  first,  the 


1  I! 


PRACTICAL  MA  THEM  A  TICS 


sharp  V '-thread,  or  common  V -thread;  and  second,  the  United 
States  standard  screw  thread.  Other  kinds  will  be  considered 
on  page  437. 

115.  Sharp  V-thread. — The  sharp  V-thread,  or  common 
V-thread,  is  a  thread  having  its  sides  at  an  angle  of  60°  to 
each  other  and  perfectly  sharp  at  the  top  and  bottom. 

The  objections  urged  against  this  thread  are,  first,  that  the 
top,  being  so  sharp,  is  injured  by  the  slightest  accident; 
and  second,  that  in  the  use  of  taps  and  dies  the  fine  sharp 
edge  is  quickly  lost,  thus  causing  constant  variation  in  fitting. 


V-Thread 


FIG.  76. 


FIG.  77. 


The  common  V-thread  with  a  pitch  of  1  in.  has  a  depth 
equal  to  the  altitude  of  an  equilateral  triangle  1  in.  on  a  side. 
Hence  its  depth  =  %\/3  =  0.866  in.  The  root  diameter  equals 
the  diameter  of  the  bolt  less  twice  the  depth  of  the  thread. 
We  have  then  the  following  formulas: 


,     «fiAA       0.866 
a  =  0.866p  =  — — > 


[14]  Dj  =  D-2d  =  D- 


1.732 


where  p  =  pitch,  rf  =  depth  of  thread,  N  =  number  of  threads 
to  the  inch,  D  =  diameter  of  bolt,  and  Di  =  root  diameter. 

116.  United  States  standard  thread.— The  United  States 
standard  screw  thread  (U.  S.  S.)  has  its  sides  also  at  an  angle 
of  60°  to  each  other  but  has  its  top  cut  off  to  the  extent  of  $ 
its  depth  and  the  same  amount  filled  in  at  its  bottom,  thus 
making  the  depth  f  that  of  the  common  V-thread  of  the  same 
pitch.  The  distance  /  on  the  flat  is  J-  its  pitch. 


TRIANGLES 


145 


This  thread  is  not  so  easily  injured,  the  taps  and  dies 
retain  their  size  longer,  and  bolts  and  screws  having  this 
thread  are  stronger  and  have  a  better  appearance. 


ILS.Staniiard 


FIG.  78. 

For  the  U.  S.  S.  screw  thread  we  have  the  following  formulas : 


_^ 

~N' 


0.6495 


N 


1.299 


[15]  D!  =  D-2d  =  D- 


The  blank  nut  for  a  bolt  is  drilled  with  a  tap  drill  the  same 
size  as  the  root  diameter  of  the  screw  or  bolt,  or  very  nearly 
that  size. 

Table  V  can  be  used  in  computations  connected  with  screw 
threads. 

EXERCISES  37 

1.  Find  the  depths  of  common  V-threads  of  the  following  number  of 
threads  to  an  inch:  10,  20,  5,  8,  40,  13. 

Ans.  0.0866  in.;  0.0433  in.;  0.1732  in.;  0.1083  in.;  0.0217  in.;  0.0666  in. 

2.  Find  the  depths  of  U.  S.  S.  threads  of  same  pitches  as  in  exercise  1. 
Ans.  0.0649  in. ;  0.0325  in. ;  0.1299  in. ;  0.0812  in. ;  0.0162  in. ;  0.0500  in. 


146  PRACTICAL  MATHEMATICS 

3.  Find  the  root  diameter  of  a  screw  of  outer  diameter  i  in.  and  14 
sharp  V-threads  to  the  inch.  Ans.  0.3763  in. 

4.  Show  that  the  depth  of  any  sharp  V-thread  in  inches  is  ^-\/3  divided 
by  the  number  of  threads  per  inch,  or,  what  is  the  same  thing,  0.866 
divided  by  the  number  of  threads  per  inch. 

6.  Check  the  depth  of  the  sharp  V-thread  for  five  of  the  sizes  given 
in  Table  V. 

6.  Using  the  formula,  find  the  root  diameter  of  the  following  common 
V-threads:  1  in.   diameter  and  20-pitch,    f  in.-16,    §  in.-13,   1$  in.-6, 
2ft  in.-4J,  3J  in.-3J. 

Ans.  0.1634  in.;  0.2668  in.;  0.3668  in.;  1.21 13  in.;  1.9276  in.;  2.9671  in. 

7.  Find  the  size  of  tap  drill  for  a  fs-in.  12-pitch  sharp  V-thread  nut. 

Ans.  0.4182  in. 

8.  What  is  the  tap  drill  size  for  the  nut  of  a  ft-in.  20-pitch  common 
double-threaded  nut?  Ans.  0.4759  in. 

9.  Find  the  depth  of  the  U.  S.  S.  screw  thread  when  there  are  15  threads 
to  the  inch.     16  threads.  Ans.  0.0433  in.;  0.0406  in. 

10.  Show  that  the  depth  in  inches  of  any  U.  S.  S.  is  |\/3  =  0.6495 
divided  by  the  number  of  threads  per  inch. 

11.  Using  the  diameter  of  the  screw,  check  the  depth  of  the  U.  S.  S. 
thread  for  five  of  the  sizes  given  in  Table  V. 

12.  Using  the  formula,  find  the  root  diameter  of  the  following  U.  S.  S. 
threads:  f  in.- 11,  1  in.-8,  If  in.-5. 

Ans.  0.507  in.;  0.838  in.;  1.490  in. 

13.  Check  the  root  diameters  for  five  sizes  of  screws  found  in  Table  V. 


CHAPTER  XII 
CIRCLES 

117.  The  importance  of  a  geometrical  form  in  the  study  of 
practical  mathematics  is  determined,  to  a  great  extent,  by  the 
frequency  of  its  occurrence  in  the  applications.     The  circle 
occurs  often,  perhaps  more  frequently  than  any  other  geomet- 
ric form  in  applied  mathematics.     Wires,  tanks,  pipes,  steam 
boilers,  pillars,  etc.,  involve  the  circle.     In  the  present  chapter 
will  be  considered  the  more  useful  facts  about  the  circle,  and 
some  of  their  applications.     Again,   the   student  is   recom- 
mended to  select  those  parts  that  are  most  closely  connected 
with  his  work  or  interests. 

118.  Definitions. — A  circle  is  a  plane  figure  bounded  by  a 
curved  line  every  point  of  which  is  the  same  distance  from 
another  point,  called  the  center. 

The  curved  line  is  called  the  circumfer- 
ence. 

A  line  drawn  through  the  center  and 
terminating  in  the  circumference  is  called 
a  diameter. 

A  line  drawn  from  the  center  to  the  cir- 

c  n    J  J-  FlG-   80- 

cumierence  is  called  a  radius. 

Any  part  of  the  circumference  is  called  an  arc. 
In  Fig.  80,  BC  is  an  arc. 

If  the  arc  equals  ^^  of  the  circumference  it  is  1°  of  arc. 
There  are  then  360°  of  arc  in  one  circumference. 

The  straight  line  joining  the  ends  of  an  arc  is  called  a  chord. 

In  Fig.  80,  DE  is  a  chord. 

The  chord  is  said  to  subtend  its  arc. 

The  chord  DE  subtends  the  arc  DmE. 

The  area  bounded  by  an  arc  and  a  chord  is  called  a  segment. 

In  Fig.  80,  the  area  DmE  is  a  segment. 

147 


us 


PR  A  CTICAL  MA  Til  EM  A  TICS 


The  area  bounded  by  two  radii  and  an  arc  is  called  a  sector. 
In  Fig.  80,  the  area  BOC  is  a  sector. 

Circles  are  said  to  be  concentric  when  they  have  a  common 
center  as  in  Fig.  81. 

D 


Fio.  81. 


Fio.  83. 


A  polygon  is  inscribed  in  a  circle  when  it  is  inside  the  circle 
and  has  its  vertices  on  the  circumference.  The  circle  is  then 
circumscribed  about  the  polygon. 

The  polygon  ABCDE  in  Fig.  82  is  inscribed  in  the  circle  O. 
A  line  is  tangent  to  a  circumference  when  it  touches  but 
does  not  cut  through  the  circumference. 

In  Fig.  83,  AT  is  tangent  to  the  circle  O  at  the  point  T. 

The  point  T  where  it  touches  the  circle  is  called  the  point 
of  tangency. 


A  polygon  is  circumscribed  about  a  circle,  or  the  circle  is 
inscribed  in  a  polygon,  when  the  sides  of  the  polygon  are  all 
tangent  to  the  circle. 

In  Fig.  84,  the  polygon  ABCDE  is  circumscribed  about  the  circle  0. 

A  central  angle  is  an  angle  with  its  vertex  at  the  center  of 
the  circle. 

In  Fig.  85,  angle  AOR  is  a  central  angle. 


CIRCLES 


149 


An  inscribed  angle  is  an  angle  with  its  vertex  on  the  cir- 
cumference of  the  circle. 

In  Fig.  85,  angle  CDE  is  an  inscribed  angle. 

An  inscribed  or  a  central  angle  is  said  to  intercept  the  arc 
between  its  sides. 

The  sides  of  the  angle  AOB  intercept  the  arc  AB,  and  the  sides  of  the 
angle  CDE  intercept  the  arc  CE. 

119.  Properties  of  the  circle. — The  student  should  become 
familiar  with  the  following  properties,  and  satisfy  himself 
that  they  are  true  by  actual  drawings  and  measurements. 

(1)  In  the  same  circle  or  in  equal  circles,  chords  that  are  the 
same  distance  from  the  center  are  equal. 

(2)  A  radius,  drawn  to  the  center  of  a  chord,  is  perpendicular 
to  the  chord  and  bisects  the  arc  which  the  chord  subtends. 

In  Fig.  86,  the  radius  OC  is  drawn  through  the  center  of  the  chord  AB. 
It  is  perpendicular  to  AB,  and  makes  arc  A (7  =  arc  CB.  This  appears 
true  by  measuring  the  parts  in  the  drawing. 

A  J3^  A- 

C 


FIG.   86. 


FIG.  88. 


(3)  The  angle  at  the  center,  that  intercepts  an  arc,  is  double 
the  inscribed  angle  that  intercepts  the  same  arc. 

In  Fig.  87,  the  central  angle  ,40C  =  600,  and  the  inscribed  angle  ABC 
is  measured  and  found  to  equal  30°. 

The  student  should  draw  several  such  figures  and  measure  the  angles. 
(See  Art.  137  for  measuring  angles.) 

(4)  The  central  angle  has  as  many  degrees  in  it  as  there  are 
in  the  arc  its  sides  intercept;  and  it  is  said  that  the  central  angle 
is  measured  by  the  arc  its  sides  intercept. 

This  is  so  because  there  are  4  right  angles  or  360°  in  the  angles  at  the 
center,  and  the  circumference  also  contains  360°. 

(5)  The  inscribed  angle  has  one-half  as  many  degrees  as  the 
arc  its  sides  intercept,  and  hence  the  inscribed  angle  is  measured 
by  one-half  the  arc  its  sides  intercept. 


150 


I'ltA  ('TIC  A  L  MA  Til  KM  A  TICS 


(0)  A  radius  drawn  to  the  point  of  contact  of  a  tangent  is 
perpendicular  to  the  tangent. 

In  Fig.  88  ,OP  is  drawn  to  the  point  of  contact  of  the  tangent  Ali.  The 
angles  oan  be  measured  and  found  to  be  right  angles.  Hence  the 
radius  is  perpendicular  to  the  tangent. 

120.  The  segment.  —  In  practical  work  it  is  often  necessary 
to  find  the  radius  of  the  circle  when  we  know  the  chord  AK 
and  the  height  of  the  segment  DC  of  Fig.  89.  If  r  stands 
for  the  radius  of  the  circle,  h  for  the  height  of  the  segment, 
and  w  for  the  length  of  the  chord,  we  have  the  following 
formulas  for  finding  r,  h,  and  w: 


[17]h=r-Vrf-(|w)2, 

=  2\/h(2r-h). 


FIQ.  89. 

For  the  derivation  of  formula  [16]  the  student  is  referred 
to  exercise  35,  page  306. 

It  should  be  noticed  that  these  formulas  are  found  by 
applying  the  principles  of  the  right  triangle. 

Example.  If  the  chord  of  the  segment  of  a  circle  is  5  ft. 
6  in.  and  the  height  of  the  segment  is  10  in.,  find  the  radius 
of  the  circle. 

OQ2_1_  1  M  ° 

Solution.     From  formula  [16],  r  =  =  59.45 

_  /\  lu 

.'.radius  =  59.45  in. 

The  method  given  in  this  article  for  finding  the  radius  of 
the  circle  when  the  length  of  the  chord  and  the  height  of 
the  segment  are  known  is  used  by  street  car  trackmen  as 
follows:  A  straightedge  10  ft.  long  is  laid  against  the  rail 


CIRCLES  151 

on  the  inside  of  the  curve.  The  distance  from  the  center  of 
the  straightedge  to  the  rail  is  measured.  This  is  the  height 
of  the  segment,  or,  as  it  is  usually  called,  the  "middle  ordinate." 
The  radius  can  now  be  found  by  the  formula  given. 

For  the  use  of  the  practical  man,  tables  are  arranged  which 
give  the  radius  corresponding  to  any  "middle  ordinate"  for 
the  10-ft.  chord. 

121.  Relations  between  the  diameter,  radius,  and  cir- 
cumference.— If  the  diameter  and  the  circumference  of  a 
circle  be  measured,  and  the  length  of  the  circumference  be 
divided  by  the  length  of  the  diameter  the  result  will  be  nearly 
3f.  This  value  is  the  ratio  of  the  circumference  to  the  di- 
ameter of  a  circle,  and  cannot  be  expressed  exactly  in  figures. 
In  mathematics  the  ratio  is  represented  by  the  Greek  letter 
TT  (pi).  The  exact  numerical  value  of  TT  cannot  be  expressed. 
The  value  to  four  decimal  places  is  3.1416.  (See  Table  II.) 

Because  of  this  relation,  if  the  diameter,  the  radius,  or  the 
circumference  is  known,  the  other  two  can  be  found. 

RULE.  The  radius  equals  one-half  the  diameter,  or  the  diam- 
eter equals  twice  the  radius. 

The  circumference  equals  the  diameter  times  3.1416. 

The  diameter  equals  the  circumference  divided  by  3.1416. 

If  r  stands  for  the  radius,  d  for  the  diameter,  and  C  for  the 
circumference,  the  rules  are  stated  in  the  following  formulas: 

[19]   C=7rd. 

[20]  d  =  C-^7r. 
[21]  C  =  27rr. 
[22]  2r  =  C-^7T. 

122.  Area  of  the  circle  — The  method  of  finding  the  area 
of  a  circle  when  the  radius,  diameter,  or  circumference  is  given 
is  established  in  geometry.  The  following  will  show  the 
reasonableness  of  the  rules. 

In  Fig.  90,  suppose  the  half  of  the  circle  AnB  is  cut  as  indi- 
cated from  the  center  nearly  to  the  circumference,  and  then 
spread  out  as  in  (6).  The  length  AnB  of  (6)  is  the  half  cir- 
cumference. Let  the  other  half  of  the  circle  be  cut  in  the 
same  manner  and  fitted  into  the  first  half.  It  is  evident  that, 
if  we  make  the  number  of  the  cuts  large,  the  figure  formed  will 


lf>2  PRACTICAL  MATHEMATICS 

be  approximately  a  rectangle  whose  length  is  equal  to  one- 
half  the  circumference,  and  whose  width  is  equal  to  the  radius. 
We  then  have  the  following: 


n  B 

(b) 


Fia.  90. 


RULE.  To  find  the  area  of  a  circle,  multiply  one-half  the 
circumference  by  the  radius. 

This  may  be  put  in  either  of  the  following  forms  which  are 
usually  more  convenient  to  use. 

RULE.  The  area  of  a  circle  equals  IT  times  the  square  of  the 
radius;  or  the  area  of  a  circle  equals  one-fourth  of  IT  times  the 
square  of  the  diameter. 

If  A  stands  for  the  area,  C  for  the  circumference,  d  for  the 
diameter,  and  r  for  the  radius,  these  rules  are  stated  in  the 
following  formulas: 

[23]  A  =  £Cr. 

[24]  A  =  7rr2  =  3.1416 Xr2. 

[25]  A=  i7rd2  =  0.7854  Xd2. 

From  formula  [24],  if  the  area  of  the  circle  is  given,  the 
radius  equals  the  square  root  of  the  quotient  when  the  area 
is  divided  by  r.  Or,  in  a  formula, 

[26]  r  = 
From  formula  [25],  we  get 


[27]  d  =  VA  -r-  ATT  =  VA  -^  0.7854. 

Example.     Find  the  radius  of  a  circle  whose  area  is  28  ft.2 
Solution.     Using  formula  [26]  and  putting  in  the  numbers, 


=  V28-f-3.14J6  =  V8.9126  =  2.985. 
/.  radius  =  2.986  ft. 


CIRCLES 


153 


123.  Area  of  a  ring. — In  the  ring,  which  is  the  area  between 
the  circumferences  of  two  concentric  circles,  the  area  can  be 
found  by  subtracting  the  area  of  the  small  circle  from  the  area 
of  the  large  circle. 


FIG.  91. 


If  A  and  a,  R  and  r  stand  for  the  areas  and  the  radii  respec- 
tively of  the  two  circles,  and  Ar  for  the  area  of  the  ring,  then 

[28]  Ar=A-a=rR2-7rr2=7r(R2-r2)=7r(R+r)(R-r). 

This  last  is  a  very  convenient  formula  to  use.  It  may  be 
stated  in  words  as  follows: 

RULE.  To  find  the  area  of  a  ring,  multiply  the  product  of 
the  sum  and  the  difference  of  the  two  radii  by  TT. 

Example.  Find  the  area  of  a  ring  of  inner  diameter  8  in. 
and  outer  diameter  12  in. 

Solution.     Using  formula  [28]  and  putting  in  the  numbers, 

Ar  =  3.1416(6+4)(6-4) 

=  3.1416X10X2  =  62.832. 
.'.area  =  62. 832  in.2 

It  should  be  noted  that  the  rule  holds  even  though  the 
circles  are  not  concentric,  that  is,  the  circles  may  be  as  in 
Fig.  91(6). 

124.  Area  of  a  sector. — The  area  of  a  sector  of  a  circle  is 
equal  to  that  fractional  part  of  the  area  of  the  whole  circle 
that  the  angle  of  the  sector  is  of  360°.  Thus,  if  the  angle  of 
the  sector  is  90°,  the  area  of  the  sector  is  -^VV  of  the  area  of 
the  circle. 

Example.  Find  the  area  of  a  sector  of  60°  in  a  circle  of 
radius  10  in. 


154 


PRACTICAL  MA  THEM  A  TICS 


Solution.  In  Fig.  92,  the  sector  AOB  has  an  angle  of  60°. 
Its  area  equals  6c%X7rr2.  If  the  radius  is  10  in.,  the  area 
of  the  sector  is 

B  A=iX3.1416X102  =  52.36in.2  Ann. 

If  6  (the  Greek  letter  theta)  stands 
for  the  number  of  degrees  in  the  angle 
of  the  sector,  and  the  other  letters  the 
same  as  before,  the  area  of  the  sector 
is  given  by  the  formula, 


Fio.  92. 

125.  Area  of  a  segment.  —  In  Fig.  92, 

it  is  evident  that  the  area  of  the  segment  ABD  equals  the  area 
of  the  sector  AOB  minus  the  area  of  the  triangle  A  OB. 
Since  it  requires  a  knowledge  of  trigonometry  to  find  the 
area  of  a  triangle  when  we  have  only  two  sides  and  an  angle, 
or  to  find  the  area  of  a  sector  when  the  angle  is  unknown, 
we  cannot  usually  find  the  area  of  a  segment  by  geometry. 
(See  page  434.) 

If  the  angle  and  the  lines  in  the  segment  are  measured, 
the  area  of,  first,  the  sector,  and  second,  the  triangle,  can  be 
found.  The  difference  be- 
tween these  areas  is  the 
area  of  thp  segment. 

Example  1.  Find  the 
area  of  a  segment  in  a 
circle  of  radius  11  j  in.  and 
subtending  an  angle  at  the 
center  of  105°. 

Solution.  The  dimen- 
sions are  as  shown  in  Fig. 
93,  where  the  parts  are 
constructed  accurately  to  scale  and  measured. 

The  area  of  sector  OADB  ***$%%  of  the  area  of  the  circle. 

.'.area  of  sector=  18$X3.1416X(11})2  =  115.97  in.-' 

Area  of  triangle  OAB  =  \  X  18  X  6  J  =  60.75  in.2 

Area  of  segment  =  area  of  sector  —  area  of  triangle 
=  115.97  in.2  -(50.75  in.  -  =  55.22  in.2     Am. 


FIG.  93. 


CIRCLES 


155 


Many  approximate  rules  are  given  to  find  the  area  of  a 
segment.  Perhaps  the  following  are  as  good  as  any. 

[30(a)]  A  =  fhw+~> 

2w 

[30(b)]  A  =  |h2^-0.608. 

In  these  rules  A  is  the  area  of  the  segment,  h  the  height,  and 
w  the  width,  while  r  is  the  radius  of  the  circle  to  which  the 
segment  belongs. 

If  the  height  of  the  segment  is  less  than  one-tenth  the  radius 
of  the  circle,  formula  [30 (a)]  may  be  shortened  to  A  =  %hw. 

Steam  engineers  often  wish  to  find 
the  area  of  a  segment  when  the  height 
is  large  compared  with  the  radius,  say, 
two-thirds  of  the  radius.  They  then 
proceed  as  follows:  In  Fig.  94,  let  it 
be  required  to  find  the  area  of  the 
segment  CnD.  Find  the  area  of  the 
half  circle  AnB,  and  then  the  area  of 
the  part  ACDB  considered  as  a  rectan- 
gle. The  area  of  the  segment  is  roughly  the  difference  be- 
tween these. 

Example  2.  Find  the  area  of  the  segment  whose  chord  is 
10  ft.  and  height  1.5  ft. 

Solution.     By  formula  [30(a)]. 

1     K3 

=  10.17-  ft.2 


FIG.   94. 


By  formula  [30  (b)],  first  finding  r  by  formula  [16], 


2X1.5 


-0.608  =  10.175  ft,2 


126.  The  ellipse.  —  The  ellipse  is  a  figure  bounded  by  a 
curved  line  such  that  the  sum  of  the  distances  of  any  point  in 
the  boundary  from  two  fixed  points  is  constant,  that  is,  always 
the  same. 


Thus,  in  Fig.  95,  any  point  P  has  the  distances  PF  -\-PF'  equal  to  the 
distances  P'F+P'F',  drawn  from  any  other  point  P'. 


156 


PRACTICAL  MATHEMATICS 


F  and  F'  are  the  two  fixed  points  and  arc  called  the  foci 
(singular  focus).  The  point  0  is  the  center  of  the  ellipse. 
NA  is  the  major  axis,  and  MB  is  the  minor  axis.  OA  and  OB 
are  the  semi-axes. 

If  a  stands  for  OA  and  b  for 
OB,  it  has  been  proved  that 
the  area  of  the  ellipse  is  given 
by  the  formula, 

[31]  A=7rab. 

Example  1.     Find  the  area 
Fio.  95.  of  an  ellipse  whose  two  axes 

are  30  ft.  and  26  ft.  respectively. 

Solution.     Using  formula  [31]  and  putting  in  the  values, 

4=3.1416X15X13  =  612.612  ft.2 

While  the  area  of  an  ellipse  is  easily  found  when  the  major 
and  minor  axes  are  given,  the  circumference,  or  perimeter,  of 
the  ellipse  is  determined  with  difficulty.  Various  approximate 
formulas  are  given  for  finding  the  circumference  of  an  ellipse. 
If  the  ellipse  is  very  nearly  the  shape  of  a  circle,  that  is,  if 
the  major  and  minor  axes  are  nearly  equal,  then 

[S2(a>]  P=T(a+b), 

where  P  is  the  perimeter  or  circumference,  a  the  semi-major 
axis,  and  b  the  semi-minor  axis. 

When  the  ellipse  differs  considerably  from  a  circle,  that  is, 
when  there  is  considerable  difference  between  the  major  and 
minor  axes,  either  of  the  following  rules  may  be  used  to  good 
advantage  : 

[32(b)]  P=T[f(a+b)-yab], 
[32(c)]  P  =  7rV2(a2+b2). 

The  exact  formula  derived  by  the  methods  of  higher  mathe- 
matics may  be  stated  in  the  following  form  : 


where  e  —  — 


0, 


This  formula  is  not  given  with  the  inten- 


tion  that  it  should  be  used,  as  the  computation  required  is 
considerable. 

Example  2.     Find  the  circumference  of  an  ellipse  whose 
major  axis  is  18  in.  and  whose  minor  axis  is  6  in. 


CIRCLES  157 

Solution.     Here  a  =  9  and  6  =  3. 
By  [32 (a)],  P  =  3.1416(9+3)  =37.699  in. 
By  [32(b)],  P  =  3. 1416[|(9+3) - \/9X3]  =  40.225  in. 
By  [32 (c)],  P  =  3.1416\/2(92+32)  =42.149  in. 
Formula  [32 (b)]  is  the  best  to  use  when  the  two  axes  are  not 
very  nearly  equal. 

EXERCISES  38 

1.  Using  d  for  diameter,  C  for  circumference,  r  for  radius,  and  A  for 
area  of  a  circle,  given  the  values  in  the  first  column  to  find  those  in 
the  last  two. 

(a)  d  =  75  ft.  C  =235.62  ft.;  A  =4417.86  ft,2 

(b)  r=23ft.  C  =  144.51  ft.;  A  =1661.90  ft.2 

(c)  d  =  34.6  in.  C  =  108. 70  in.;  A  =940.25  in.2 

(d)  d  =  24.5  in.  C  =  76.97  in.;  A  =471. 44  in.2 

(e)  C  =  86.08  in.  d  =  27 A  in.;  A  =589.65  in.2 

(f)  (7  =  158.02  in.  d  =  50.3  in.;  A  =  1987.13  in.2 

(g)  A  =  1452.20  ft.2  d  =  43ft.;  (7  =  135.09  ft. 
(h)  A  =27171. 6  ft.2  r=93ft.;  (7  =  584.34  ft. 

2.  Find  the  area  of  the  cross  section  of  a  half-inch  rod. 

Ans.  0.196+  in. » 

By  the  cross  section  is  meant  the  area  of  the  end  of  the  rod  when  cut 
square  off. 

3.  Find  the  area  of  the  ring  enclosed  between  two  circles,  the  outer 
9  in.  and  the  inner  8  in.  in  diameter.  Ans.  13.352  in.2 

4.  The  inner  and  outer  diameters  of  a  ring  are  9?  and  10  in.  respec- 
tively.    Find  the  area  of  the  ring?  Ans.  7.658  in.z 

5.  Find  the  area  of  the  ring  in  the  cross  section  of  a  water  main  40  in. 
in  external  diameter,  if  the  iron  is  1  in.  thnk  in  the  shell. 

Ans.  122.52+  in.2 

6.  In  an  elliptical  garden  the  longest  diameter  is  36  ft.  and  the  shortest 
22  ft.     Find  the  area  of  the  garden.  Ans.  C22.04-  ft.2 

7.  In  a  steel  plate  3  ft.  by  2^  ft.  are  26  round  holes,  each  If  in.  in  diam- 
eter.    Find  the  area  of  steel  remaining.  Ans.  1017.46+  in.2 

8.  At  the  center  of  one  side  of  a  barn  40  ft.  on  a  side  a  horse  is  tied 
by  a  rope  70  ft.  in  length.     Find  the  area  he  can  graze  over  in  square 
rods.  Ans.  43.27+  rd.2 

Suggestion.  When  the  figure  is  drawn  it  is  seen  that  the  horse  can 
graze  over  a  half-circle  having  a  radius  of  70  ft.,  two  quarter-circles  having 
a  radius  of  50  ft.,  and  two  quarter-circles  having  a  radius  of  10  ft, 

9.  A  6-in.  water  pipe  can  carry  how  many  times  as  much  as  an  inch 
pipe? 

Solution.     Area  of  6-in.  pipe  =0.7854X62  in.2 
Area  of  1-in.  pipe  =0.7854  XI2  in.2 
Area  of  6-in.  pipe      0.7854X62     62 

1-1 —  .— .    — __ —   —     __     -IK         4  99£ 

Area  of  1-in.  pipe     0.7854 XI2     I2 


158 


PRACTICAL  MA  THEM  A  TICS 


The  quotient  or  the  ratio  of  the  areas  of  two  circles  can  always  be  found 
by  dividing  the  square  of  one  diameter  by  the  square  of  the  other.  The 
radii  may  be  used  instead  of  the  diameters. 

The  above  is  simply  the  principle  that  similar  areas  are  in  the  same 
ratio  as  the  squares  of  their  like  dimensions,  applied  to  circles. 

10.  In  putting  up  blower  pipes,  two  circular  pipes  11  in.  and  14  in. 
in  diameter  respectively  join  and  continue  as  a  rectangular  pipe  14  in. 
in  width.     Find  the  length  of  the  cross-section  of  the  rectangular  pipe. 

Ana.  17.78+  in. 

11.  How  many  times  the  area  of  the  cross  section  of  a  ^-in.  wire  is 
a  half-inch  wire?  Ana.  64. 

12.  If  an  inch  pipe  will  empty  2  barrels  in  15  minutes,  how  many 
barrels  will  an  8-in.  pipe  empty  in  24  hours?     (Make  no  allowance  for 
friction.)  Ana.  12,288. 

13.  How  many  3-in.  steam  pipes  could  open  off  from  an  18-in.  steam 
pipe?  Ana.  36. 

14.  The  diameter  of  the  safety  valve  in  a  boiler  is  3  in.     Find  the  total 
pressure  tending  to  raise  the  valve  when  the  pressure  of  the  steam  is 
120  Ib.  per  square  inch.  Ans.  848.23  Ib. 

15.  If  the  diameter  of  a  piston  is  30  in.,  find  the  total  pressure  on  the 
piston  when  the  pressure  of  steam  is  100  Ib.  per  square  inch. 

Ans.  70,686  Ib. 

16.  A  circular  sheet  of  steel  2  ft.  in  diameter  increases  in  diameter  by 
zJo  when  the  temperature  is  increased  by  a  certain  amount,     (a)  Find 
the  increase  in  the  area  of  the  sheet.     (6)  Find  the  per  cent  of  increase 
in  area.  Ans.  (a)  0.03149  ft.*;  (b)  1%. 

17.  How  many  No.  20  B.  and  S.  copper  wires  will  have  the  same  cross 
section  area  as  one  No.  00?     (See  Table  VII.)  Ans.  130.3 -. 


Fiu.  97. 


18.  A  hot-air  pipe  9  in.  in  diameter  passes  into  a  boot  as  shown  in 
Fig.  96,  and  a  rectangular  pipe  of  same  capacity  passes  upward  from  the 
boot.     If  the  rectangular  pipe  is  4  in.  wide,  find  its  length  in  cross  section. 

Ana.  15.9+  in. 

19.  Given  two  joining  pipes  12  in.  and  8  in.  in  diameter  respectively, 
to  find  the  diameter  x  of  the  continuation  which  has  the  same  area. 
(See  Fig.  97.)  Ans.  14.4+  in. 

20.  Given  an  elliptical  pipe  of  longest  and  shortest  axes  16  in.  and  1C 


CIRCLES 


159 


in.  respectively,  to  find  the  diameter  of  the  circular  pipe  having  the  same 
area  of  cross  section.  Ans.  12.65—  in. 

21.  Show  by  means  of  the  carpenter's  square  how  to  find  the  diameter 
of  a  circle  having  the  same  area  as  the  sum  of  the  areas  of  two  given 
circles. 

Discussion.  Suppose  we  take  two  circles  6  in.  and  8  in.  in  diameter 
respectively.  Lay  off  on  one  arm  of  the  carpenter's  square,  as  shown  in 
Fig.  98,  the  diameter  of  the  6-in. 
circle  and  on  the  other  arm  the 
diameter  of  the  8-in.  circle.  The 
line  joining  the  ends  of  these,  or  the 
hypotenuse  of  the  right  triangle,  is 
the  diameter  of  the  circle  having 
the  same  area  as  the  sum  of  the 
areas  of  the  two  given  circles. 

This  is  seen  to  be  true  in  this 
particular  case  as  follows: 

Area  of  6-in.  circle  =62X0.7854  in.2 

Areaof  8-in.  circle  =82X 0.7854 in.2 

Sum  of  areas  =(62+82)X  0.7854  in.2 
But  62+82  =  102,  and  if  6  in.  and  8 
in.  are  respectively  the  altitude  and 
base  of  a  right  triangle  then  10  in.  is 
the  hypotenuse. 

Hence  the  area  of  the  circle  equal 
to  the  sum  is  102X0.7854  in.2     The  diameter  of  this  circle  is  evidently  10 
in.,  which  is  the  hypotenuse  of  the  right  triangle  as  drawn  in  the  figure. 

A  similar  discussion  would  apply  to  any  two  circles. 

22.  Show  by  means  of  the  carpenter's  square  how  to  find  the  diameter 
of  a  circle  having  the  same  area  as  the  sum  of  the  areas  of  any  number  of 
given  circles. 

23.  If  the  drive  wheels  of  a  locomotive  are  66  in.  in  diameter,  find  the 
number  of  revolutions  per  minute  to  go  40  miles  per  hour. 

Am.  203.7 +  . 


FIG.  98. 


40X5280X12 


;=  203.7. 


60X66X3.1416 

Solution.  It  is  usual  to  work  such  problems  as  this  by  cancellation. 
Above  the  line  are  the  numbers  which  give  the  inches  in  40  miles.  Below 
the  line  is  60,  which  we  divide  by  to  get  the  number  of  inches  the  train 
goes  in  1  minute;  and  66X3.1416,  which  is  the  number  of  inches  in  the 
circumference  of  the  wheel. 

24.  Supposing  that  the  driving  wheels  of  a  locomotive  are  16  ft.  in 
circumference,  what  number  of  revolutions  must  they  make  per  minute 
so  that  the  locomotive  may  attain  a  speed  of  60  miles  per  hour? 

Ans.  330. 

25.  A  locomotive  wheel  5  ft.  in  diameter  made  10,000  revolutions  in  a 


100  1'RACTICAL  MATHEMATICS 

distance  of  24  miles.     What  distance  was  lost  due  to  the  slipping  of  the 
wheels?  Ana.  5}  miles. 

26.  If  an  arc  of  a  circle  is  equal  in  length  to  the  radius,  what  is  the  value 
of  the  central  angle  which  it  measures?  Ann.  57.2958°  —  . 

Solution.  Since  2ir  times  the  radius  equals  the  circumference,  and  the 
entire  circumference  measures  an  angle  of  360°  at  the  center  of  the  circle, 
the  number  of  degrees  =  360  -^2ir  =57.2958—  . 

27.  The  radii  of  two  circles  are  2  ft.  and  4  ft.     The  area  of  the  second 
is  how  many  times  the  first?  Ans.  4. 

28.  The  length  of  the  circumference  of  a  circle  is  132  ft.     Calculate 
the  length  of  the  diameter,  the  length  of  an  arc  of  40°,  and  the  area  of  a 
sector  of  80°.  Ans.  Arc  14.67-  ft.;  dia.  42.017-  ft.;  area  308.  1+  ft.1  • 

29.  Find  the  weight  of  the  iron  hoops  on  a  tank  15  ft.  in  diameter, 
there  being  16  hoops  weighing  3  Ib.  per  linear  foot.   Ans.  2261.9+  Ib. 

B  30.  A  regular  hexagon,  the  perimeter  of 
which  is  42  ft.,  is  inscribed  in  a  circle.  Find 
the  area  of  the  circle.  Ans.  153.9+  ft.2 

31.  What   is    the   waste   in   cutting   the 
largest  possible  circular  plate  from  a  piece 
of  sheet  steel  17  in.  by  20  in.? 

Ans.  113.02  in.2 

32.  Four   of   the   largest   possible   equal 
sized  pipes  are  enclosed  in  a  box  of  square 

cross  section  18  in.  on  an  edge.     What  part  of  the  space  do  the  pipes 
occupy?  Ans.  0.7854. 

33.  Find  size  of  the  box  to  enclose  five  6-in.  pipes,  placed  as  in  Fig.  99, 
and  find  the  part  the  area  of  the  pipes  is  of  the  area  of  the  box. 

Ans.  Box  12  in.  by  16.392  in.;  part  occupied  0.7187+. 
Solution.     AC  =  12  in.,  AB  =  2DM. 

DM  =  DN+NM,  but  DN  =3  in.  and 


.*.  DM  =  3  in.  +5.196  in.  =8.196  in. 
A  AB  =  2X8.196  in.  =  16.392  in. 
/.  area  =  12  X  16.392  =  196.704  in.2 
Area  of  5  circles  =5X0.  7854  X62  =  141.372  in.2 
Part  occupied  by  pipes  =  141.372-^-196.704  =0.7187+. 
34.  If  the  diameter  of  a  circle  is  3  in.,  what  is  the  length  of  an  arc  of 
80°?  Ans.  2.0944  in. 

36.  The  minute  hand  of  a  tower  clock  is  6  ft.  long.     What  distance 
will  the  extremity  move  over  in  36  minutes?          Ans.  22  ft.  7.4+  in. 

36.  The  maximum  circumferential  velocity  of  cast-iron  flywheels  is 
80  ft.  per  second.     Find  the  maximum  number  of  revolutions  per  minute 
for  a  cast-iron  flywheel  8  ft.  in  diameter.  Ans.  191  nearly. 

37.  An  emery  wheel  may  have  a  circumferential  velocity  of  5500  ft. 
per  minute.     Find  the  number  of  revolutions  per  second  an  emery  wheel 
9  in.  in  diameter  may  make.  Ans.  39  nearly. 


CIRCLES  161 

38.  The  peripheral  speed  of  a  grindstone  of  strong  grain  should  not 
exceed  47  ft.  per  second.     Find  the  number  of  revolutions  per  minute  a 
grindstone  3  ft.  in  diameter  may  turn.  Ans.  299  nearly. 

39.  The  area  of  a  square  is  49  sq.  ft.     Find  the  length  of  the  circum- 
ference and  the  area  of  the  circle  inscribed  in  this  square. 

Ans.  21.99+  ft.;  38.48+  ft.2 

40.  Find  the  size  of  the  largest  square  timber  which  can  be  cut  from 
a  log  24  in.  in  diameter.  Ans.  16.97+  in. 

41.  A  roller  used  in  rolling  a  lawn  is  6.5  ft.  in  circumference  and  2.5  ft. 
wide.     If  the  roller  makes  10  revolutions  in  crossing  the  lawn  once  and 
must  go  up  and  back  12  times,  what  is  the  area  of  the  lawn? 

Ans.  3900  ft.2 

42.  Three  circles  are  enclosed  in  an  equilateral  triangle.     If  the  circles 
are  10  in.  in  radius,  find  the  sides  of  the  triangle.  Ans.  54.64  in. 


A  B 

I'lG.    100. 

Suggestion.  The  circles  are  as  shown  in  Fig.  100.  The  triangle 
DEG  =  triangle  DAH.  Hence  AH  =GE  =  10  X\/3  =  17.32  in.,  #7=  the 
diameter  of  one  of  the  circles  =20  in.,  and  1C  =  AH  =  17.32  in. 

43.  Using  4000  miles  as  radius  of  earth,  find  length  in  feet  of  one  second 
of  arc  on  the  equator.  Ans.  102.4—  ft. 

Note.     1°=60  minutes  and  1  minute  =60  seconds  of  arc. 

44.  Using  4000  miles  as  radius  of  earth,  find  the  length  in  miles  of  arc 
of  1'  (a)  on  the  parallel  of  45°  north;  (b)  on  the  parallel  of  60°  north. 

Ans.   (a)  0.823-  miles;  (b)  0.582-  miles. 

Suggestion.  For  the  parallel  of  45°  north,  CB  is  the  radius.  But  CB  = 
OC  since  the  triangle  OCB  is  a  right  triangle  with  two  equal  angles.  The 
relations  are  as  shown  in  Fig.  101. 

45.  A  bicycle  is  so  geared  that  one  revolution  of  the  feet  makes  two 
revolutions  of  the  wheels  which  are  28  in.  in  diameter      How  many 
revolutions  per  minute  of  the  feet  are  necessary  to  go  at  the  rate  of  25 
miles  per  hour?     Suppose  that  the  pneumatic  tires  are  not  well  inflated, 

n 


162 


PRACTICAL  MATHEMATICS 


what  is  the  per  cent  of  loss  in  distance  made  if  the  compression  in  the 
tire  is  i  in.?  An*.  150  +  ;  3.6-%. 

Suggestion.  If  the  compression  of  the  pneumatic  tire  is  J  in.,  the  wheel 
acts  as  if  it  were  J  in.  less  in  radius. 

46.  What  is  the  per  cent  of  error  in  taking  4  times  CA  in  Fig.  102  as 
the  circumference  of  circle  0?  Ans.  0.66  —  %  too  large. 


B 


FIG.   101. 


Fia.   102. 


47.  In   the  same  circle,   what    is  the    per    cent    of    error  in  taking 
<l(DE  +  }AB)  as  the  circumference?  Ans.  0.65  —  %  too  small. 

48.  If  statements  in  numbers  46  and  47  gave  the  exact  length  of  the 
circumference,  what  would  be  the  value  of  v  in  each  case? 

Ans.  In  No.  46,  ir=\/T6  =  3.16228- ;  in  No.  47,  r=3.1213  +  . 

49.  Make  a  construction  as  shown  in  Fig.   103,  and  A  B  is  approxi- 
mately the  quadrant  of  the  circle.     Find  the  per  cent  it  differs  from  the 
correct  value.  Ans.  0.4  +  %  too  large. 


FIG.  104. 


Suggestion.  By  the  quadrant  of  the  circle  is  meant  the  length  of  the 
arc  BN. 

60.  Justify  the  following  rule  used  by  sheet-metal  workers,  or  show 
the  per  cent  of  error  if  it  is  not  correct:  Divide  the  radius  AO,  Fig.  104, 
into  four  equal  parts;  place  one  of  these  parts  from  A  to  C  and  another 
from  B  to  D,  the  ends  of  two  perpendicular  diameters.  Connect  C  and 
D,  which  gives  the  side  of  the  square  of  the  same  area  as  the  circle. 

Ans.  0.5  +  %  too  small. 


CIRCLES  163 

51.  The  stem  of  a  4-in.  safety  valve,  Fig.  105,  is  2j  in.  from  the  ful- 
crum.    Supposing  the  valve  will  blow  when  the  gage  reads  7  Ib.  without 
any    weight    on   the   lever    (i.e., 

that  7  Ib.  per  square  inch  on  the 
valve  overcomes  weight  of  valve 
and  lever),  at  what  pressure 
would  it  blow  with  a  weight  of 
75  Ib.  32  in.  from  the  fulcrum? 

Ans.  76.4+  Ib.  per  square  inch. 

Solution.  If  P  represents  the 
total  number  of  pounds  pressure 
on  the  valve  necessary  to  lift  the 
weight,  then 

2.75: 32  =  75: P.      From  which  FIG.  105. 

P=  872.73. 

Area  of  valve  =0.7854  X42  =  12.5664  sq.  in. 

Pressure  to  raise  weight  =  872.73 -^  12.5664  =69.4  Ib.  per  sq.  in. 

Total  pressure  =69.4  Ib.  +7  Ib.  =76.4  Ib.  per  sq.  in. 

52.  What  weight  of  ball  would  be  required  to  allow  the  valve  in  exer- 
cise 51  to  blow  off  at  80  Ib.?  Ans.  78.8+  Ib. 

53.  If  the  original  weight  of  75  Ib.  is  used,  at  what  distance  from  the 
fulcrum  should  it  be  placed  to  allow  the  valve  to  blow  off  at  80  Ib.? 

Ans.  33.6+  in. 

54.  The  following  is   an  approximate  formula  for  determining  the 
number  of  inscribed  tangent  circles  in  a  larger  circle: 


N  =  0.907  [-j- 0.94      +3.7, 

\  Of  j 

where  N  is  the  number,  D  the  diameter  of  the  enclosing  circle,  and  d 
the  diameter  of  the  inscribed  circles.  This  formula  can  be  used  to  find 
the  number  of  wires  that  can  be  put  in  a  casing  of  given  size. 

Apply  the  above  formula  and  find  how  many  wires  |  in.  in  diameter 
can  be  placed  inside  a  5-in.  pipe.  Ans.  78. 

65.  How  many  steel  balls  0.4  in.  in  diameter  can  rest  at  the  bottom 
of  a  closed  pipe  2|  in.  in  diameter?  Ans.  29. 

56.  In  a  circle  of  radius  5  ft.  there  is  a  chord  6  ft.  6  in.  in  length.     Find 
the  height  of  the  segment.  Ans.  1.200+  ft. 

57.  A  segment  of  a  circle  cut  off  by  a  chord  4  ft.  6  in.  in  length  has  a 
height  of  1  ft.  10  in.     Find  the  radius  of  the  circle.      Ans.  27.57  —  in. 

58.  Find  the  area  of  a  sector  in  a  circle  whose  radius  is  28  cm.,  if  the 
sector  contains  an  angle  of  50°  36'.  Ans.  346.19—  cm.2 

59.  Find  the  area  of  a  sector  whose  arc  is  99.58  m.  long  and  radius 
86.34m.  Ans.  4298.87-  m.2 

60.  Test  formula  [30 (a)]   by  taking  the  segment  as  half  the  circle 
60  in.  in  diameter. 

Ans.  By  [30(a)j  A  =  1425  in.2     As  f  a  circle  A  =1414—  in.2 

61.  Find  the  radius  of  a  circle  in  which  a  chord  of  10  ft.  has  a  middle 
ordinate  of  3  in,  Ans.  50  ft.  1.5  in. 


Kit 


PRACTICAL  MATHEMATICS 


62.  The  Gothic  Arch  is  formed  by  two  arcs  each  i  of  a  circle.  The 
center  of  each  circle  is  at  the  extremity  of  the  width  of  the  arch,  that  is, 
the  radius  equals  the  width.  Find  the  area  of  such  an  arch  of  radius 
6ft.  (See  Fig.  106.)  Ann.  22.11  ft.* 


Fia.  106. 


63.  In  Fig.  107,  W,  W  is  a  wall  with  a  round  corner,  of  dimensions  as 
given  from  A  to  B,  on  which  a  molding,  gutter,  or  cornice  is  to  be  placed ; 
find  the  radius  of  the  circle  of  which  arc  ANB  is  a  part.     Ans.  3  ft.  J  in. 

64.  Each  of  four  steam  engines  is  supplied  by  a  6-in.  steam  pipe. 

These  open  off  from  a  single  steam 
pipe.  Find  the  diameter  of  the  larger 
pipe  that  it  may  have  the  same 
capacity  as  the  four  6-in.  pipes. 

Ans.  12  in. 

65.  A   milling   cutter    4J    in.    in 
diameter  is  cutting  soft  steel  at  the 
rate  of  45  ft.  per  minute.     Find  the 
number  of  revolutions  per  minute. 

Ans.  38.2-. 

66.  How   many   turns   per   second 
must  a  drill  \  in.  in  diameter  make  so  that  the  outer  edge  of  the  lip  will 
have  a  cutting  speed  of  35  ft.  per  minuteV  Ans.  4.46  —  . 

67.  The  distance  between  the  center  of  the  crank  pin  C,  Fig.  109,  and 
the  center  of  the  flywheel  at  D  is  20  in.     What  is  the  length  of  the  stroke 


FIG.   108. — Milling  cutter. 


Fio.  109. 

of  the  piston?    If  the  flywheel  makes  144  R.  P.  M.,  find  the  average 
speed  of  the  piston  in  feet  per  minute.     Ans.  40  in.;  960  ft.  per  min. 
68.  The  "piston  speed  "  in  a  Corliss  engine  should  be  600  ft.  per  minute. 


CIRCLES 


165 


FIG.    110. 


How  many  revolutions  per  minute  should  be  made  by  an  engine  having 
a  20-in.  stroke?     By  an  engine  having  a  36-in.  stroke? 

Ana.  180;  100. 

69.  Which  would  occupy  the  greater  portion  of  the  square  shown  in 
Fig.  110,  the  four  small  circles  or  the  large  circle?     Ans.  Both  the  same. 

70.  The  drivers  on  a  locomotive  are  making  210  R.  P.  M.,  and  are 
76  in.  in  diameter.      Find  the  speed  of  the  loco- 
motive in  miles  per  hour  if  2%   is  allowed  for 

slipping.  Ans.  46. 53  mi.  per  hr. 

71.  In  a  Corliss  engine  the  high-pressure  cylin- 
der is  22  in.  in  diameter.      What  must  be  the 
diameter  of  the  low-pressure  cylinder  in  order  that 
it  may  have  double  the  area  of  the  high-pressure 
cylinder?  Ans.  31.1+  in. 

72.  If  the  total  pressure  on  the  piston  of  a  brake 
cylinder  is  8100  lb.,  what  is  the  diameter  of  the 

cylinder  if  the  pressure  is  60  lb.  per  square  inch?         Ans.   13.1+  in. 

73.  A   10-in  pipe  is  to  be  branched  off  into  two  equal  pipes.     What 
must  be  the  diameter  of  each  of  these  pipes  if  the  two  pipes  shall  equal 
the  area  of  the  10-in.  pipe?  Ans.  7.07+  in. 

74.  A  machine  screw  f  in.  in  diameter  has  12  sharp  V-threads  to  the 
inch.     Find  the  root  diameter.     Find  the  tensile  strength  at  50,000  lb. 
per  square  inch.     Give  answer  to  the  nearest  100  lb.         Ans.  9100  lb. 

75.  What  should  be  the  area  of  the  opening  of  a  cold-air  box  for  a  hot- 
air  furnace  in  order  to  supply  7  hot-air  pipes  9  in.  in  diameter  and  one 
pipe  14  in.  in  diameter,  if  the  area  of  the  cold-air  opening  is  -|  the  area  of 
the  hot-air  pipes?  Ans.  449.4+  sq.  in. 

76.  A  cylinder  of  a  double-acting  engine  is  26  in.  in  diameter  and  the 
length  of  the  stroke  is  30  in.     Compute  the  pressure  on  each  side  of 
the  piston  if  the  piston  rod  is  3|  in.  in  diameter  and  the  steam  pressure 
in  the  cylinder  is  150  lb.  per  square  inch.      (Use  formula  [28].) 


FIG.   111. 

77.  In  practice  piston  rings  for  a  steam  engine  are  turned  so  that  they 
are  \\%  larger  in  diameter  than  the  diameter  of  the  cylinder  barrel. 
They  then  have  a  piece  cut  out  and  are  sprung  into  place.  Find  the 
diameter  of  the  ring  for  a  cylinder  24  in.  in  diameter.  Find  the  length 
of  the  piece  to  be  cut  out  if  when  sprung  into  place  it  has  a  clearance  of 
i^  in.  between  ends.  Give  dimensions  to  the  nearest  32nd  of  an  inch. 

Ans.  24 -f  in.;  1^  in. 


160  PRACTICAL  MATHEMATICS 

78.  Find  the  speed  of  a  belt  running  over  a  pulley  having  a  diameter 
of  22  in.  and  making  320  R.  P.  M.,  if  \%  is  allowed  for  slipping. 

Ana.  1834—  ft.  per  minute. 

79.  If  the  greatest  and  least  diameters  of  an  elliptical  manhole  arc 
2  ft.  7  in.  and  2  ft.  3  in.  respectively,  find  its  area.     Find  its  perimeter, 
using  formulas  [32 (b)]  and  [32 (c)]. 

Ans.  4.565  ft.1;  91.2+  in.  and  91.3  in. 

80.  Find  the  length  in  feet  of  the  arc  of  contact  of  a  belt  with  a  pulley, 
if  the  pulley  is  3  ft.  6  in.  in  diameter  and  the  arc  of  contact  is  210°. 

Ans.  6  ft.  5—  in. 

81.  As  in  the  preceding,  find  the  length  of  the  arc  of  contact  if  it  is 
120°  and  the  diameter  of  the  pulley  is  16  in.  Ans.  16  J  in. 


REGULAR  POLYGONS  AND  CIRCLES 

127.  It  is  often  necessary  to  determine  the  dimensions  of  a 
regular  polygon  inscribed  in  or  circumscribed  about  a  given 
circle,  or  to  determine  the  size  of  a  circle  that  can  be  inscribed 
in  or  circumscribed  about  a  given  polygon. 

Such  problems  are  readily  solved  by  trigonometry,  and  some 
of  them  may  be  solved  by  geometry.  In  either  case,  though, 
the  computation  may  be  long  and  tedious.  For  this  reason 
handbooks  give  rules  by  which  the  computations  can  be  readily 
made.  In  the  table  on  page  167  are  classified  certain  facts 
about  the  regular  polygons  named.  These  facts  can  readily 
be  applied  to  polygons  of  any  size. 

(1)  To  find  the  area  of  a  polygon  when  the  length  of  one 
side  is  given. — Multiply  the  square  of  the  side  by  a  number 
given  in  column  (3). 

This  rule  is  an  application  of  the  principle  that  similar  areas 
are  in  the  same  ratio  as  the  squares  of  their  like  dimensions. 
Show  that  this  is  the  case. 

Example.  Find  the  area  of  a  regular  pentagon  having 
sides  of  7  in. 

Area  =  72X  1.7204774  in.2  =  84.30339  in.2 

(2)  To  find  the  side  of  a  polygon  when  its  area  is  given. — 
Diinde  the  area  of  the  polygon  by  a  number  from  column   (3). 
The  square  root  of  the  quotient  is  the  required  side  of  the  polygon. 

Example.  The  area  of  an  octagon  is  4376  ft.2;  find  a  side 
of  the  polygon. 

Side  =  v/4376-*- 4.828  =  30.1  (Hi  ft. 


CIRCLES 


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11)8  PRACTICAL  MATHEMATICS 

(3)  To  find  the  radius  of  the  circumscribing  circle  when 
a  side  of  the  polygon  is  given. — Multiply  the  length  of  a  side 
by  a  number  chosen  from  column  (5). 

This  can  be  used  to  good  advantage  in  drawing  u  regular 
polygon  of  a  given  side. 

Example.     Construct  a  regular  decagon  having  sides  of  2  3  in. 

Radius  of  circumscribing  circle  =  2\ XI. 618  in.  =4.045  in. 
With  the  compasses  construct  a  circle  of  this  radius.  Then 
with  the  dividers  open  2\  in.  step  around  the  circle,  which 
should  be  divided  into  10  parts.  Connect  these  points  succes- 
sively and  the  construction  is  complete. 

(4)  To  find  the  radius  of  the  inscribed  circle  when  a  side  of 
the  polygon  is  given. — Multiply  the  length  of  a  side  by  a  number 
chosen  from  column  (6). 

(5)  To  find  the  length  of  the  side  of  a  polygon  that  can  be 
inscribed  in  a  circle   of  given  radius. — Multiply  the  given 
radius  by  a  number  chosen  from  column  (7). 

Example.  Construct  a  regular  heptagon  in  a  circle  of  3-in. 
radius. 

A  side  of  the  polygon  =3X0.8677  in.  =2.6  in. 
With  the  dividers  open  2.6  in.  step  around  the  circle,  which 
should  be  divided  into  7  equal  parts.     Connect  these  points 
successively  and  the  construction  is  complete. 

EXERCISES  39 

1.  A  round  shaft  is  3J  in.  in  diameter.     Find  the  length  of  the  side 
of  a  triangular  end  that  can  be  made  on  the  shaft.     Find  the  length  of 
the  side  of  a  square  end.     Of  a  hexagonal  end.     Of  an  octagonal  end. 
Ans.  2.81+  in.;  2.30-  in.;  If  in.;  1.24+  in. 


Triangular  Pentagonal  Hexagonal 

Fio.    112. 

2.  Find  the  diameter  of  a  circular  shaft  so  that  it  may  have  a  triangular 
end  2J  in.  on  a  side.  A  pentagonal  end  1$  in.  on  a  side.  An  octagonal 
end  1 1  in.  on  a  side.  Ans.  2.887-  in.;  2.552-  in.;  2.940-  in. 

Suggestion.     Use  rule  (3). 


CIRCLES  169 

3.  A  square  taper  reamer  is  to  be  made  which  must  ream  1|  in.  at 
the  small  end  and  If  in.  at  the  back  end.     What  must  be  the  distance 
on  the  flat  face  at  each  end?  Ans.  0.795+  in.;  1.149+  in. 

4.  What  is  the  diameter  of  the  bearing  that  can  be  turned  on  a  tri- 
angular shaft  of  side  2  in.?     On  a  hexagonal  shaft  of  side  If  in.? 

Suggestion.     Use  rule  (4).  Ans.  1.155—  in. ;  2. 815—  in. 

6.  Find  the  difference  between  the  area  of  a  circle  of  radius  5  in.  and 
the  area  of  the  inscribed  regular  triangle.  The  inscribed  regular  penta- 
gon. Hexagon.  Octagon.  Decagon. 

Ans.  46.065+  in.2;  19.10-  in.2;  13.59-  in.2;  7.94+  in.2;  5.08-  in.2 

Suggestion.  Find  the  side  of  the  inscribed  polygon  by  rule  (5)  and 
its  area  by  rule  (1). 

6.  Find  the  radius  and  area  of  the  largest  circle  that  can  be  cut  from 
a  triangle  every  side  of  which  is  4  ft.         Ans.  1.155—  ft.;  4.19—  ft.2 

7.  The  triangular  end  on  a  round  shaft  is  1.7  in.  on  a  side.     Find  the 
diameter  of  the  shaft.  Ans.  1.96+  in. 

8.  The  area  of  a  regular  hexagon  inscribed  in  a  circle  is  24\/3-     Find 
the  area  of  the  circle  and  the  length  of  the  circumference. 

Ans.  50.266-;  25.133-. 

9.  A  square  end  0.875  in.  on  a  side  must  be  milled  on  a  shaft.     What  is 
the  diameter  to  which  the  shaft  should  be  turned?         Ans.  1.237+  in. 

10.  A  pipe  10  in.  in  diameter  is  connected  to  a  hexagonal  pipe  of  the 
same  area  in  cross  section.     Find  the  edge  of  the  hexagon  of  the  cross 
section  of  the  hexagonal  pipe.  Ans.  5.50—  in. 


TURNING  AND  DRILLING 

128.  Rules. — The  cutting  speed  of  a  tool  is  the  rate  at 
which  it  passes  over  the  surface  being  cut.  This  applies  to  a 
lathe  tool  in  turning  a  piece  of  work,  such  as  a  car  axle;  or  to 
a  drill  used  in  making  holes  in  a  metal  of  any  kind. 

The  rate  at  which  the  tool  can  cut  the  metal  without  in- 
juring the  tool  depends  upon  the  material  in  the  tool,  as  well  as 
upon  the  kind  of  work  being  turned. 

Since  cutting  speeds  are  usually  given  in  feet  per  minute, 
the  rate  at  which  a  tool  is  cutting  can  be  found  by  the  following : 

RULE.  Multiply  the  circumference  of  the  piece,  or  of  the 
drill,  in  feet  by  the  number  of  revolutions  per  minute.  This 
gives  the  cutting  speed  in  feet  per  minute. 

This  applies  to  work  turned  in  a  lathe  or  to  the  drill  in  a 
drill  press. 

It  follows  from  the  above  that  the  number  of  revolutions, 
allowable  per  minute,  is  found  by  the  following: 


170  PRACTICAL  MATHEMATICS 

RULE.  Divide  the  cutting  speed  in  feet  per  minute  by  the 
circumference  of  the  work  in  feet.  This  gives  the  number  of 
revolutions  per  minute. 

129.  Feed. — The  feed  of  a  tool  is  the  sideways  motion 
given  to  the  cutting  tool.     It  is  expressed  in  one  of  the  follow- 
ing ways: 

(1)  The  feed  is  the  part  of  an  inch  that  the  tool  advances 
along  the  work  for  each  revolution  or  stroke,  as  a  feed  of  A  inch. 

(2)  The  feed  is  the  number  of  revolutions  or  strokes  neces- 
sary to  advance  the  tool  1  inch,  as  a  feed  of  20  turns  to  the  inch. 

(3)  The  feed  is  the  number  of  inches  the  tool  advances  in 
1  minute,  as  the  feed  is  f  inch  per  minute. 

Thus,  in  turning  a  car  axle,  the  shaving  may  be  J  in.  wide, 
which  means  that  the  tool  must  advance  that  distance  along 
the  axle  for  every  revolution  of  the  axle.  That  is,  it  will  take 
4  turns  of  the  axle  to  cover  1  in.  of  its  length  with  the  turning 
tool. 

130.  Cutting  Speeds. — Cutting  speeds  for  carbon  steel  tools 
should  be  about  30  ft.  per  minute  in  steel,  35  ft.  per  minute 
in  cast  iron,  and  60  to  100  ft.  per  minute  in  brass. 

The  general  rule  is  to  run  high-speed  steel  tools,  in  steel, 
about  double,  and  in  iron  about  three  times  the  speed  of  the 
carbon-steel  tool. 

The  maximum  speed  given  to  any  tool  must  be  governed  by 
the  density  and  toughness  of  the  material  being  cut,  and  by 
the  way  the  tool  "holds  up." 

The  feed  of  a  drill  should  be  from  0.004  in.  to  0.01  in.  per 
revolution. 

EXERCISES  40 

1.  In  turning  a  brass  rod  2  in.  in  diameter,  what  is  the  proper  number 
of  revolutions  per  minute  if  the  cutting  speed  for  brass  is  100  ft.  per 
minute?  Ans.  191. 

2.  In  turning  a  locomotive  wheel  78  in.  in  diameter,  what  is  the  proper 
number  of  revolutions  per  minute,  in  order  that  the  cutting  speed  may 
be  10  ft.  per  minute?  Ans.  0.49  nearly. 

3.  In  turning  a  tool-steel  arbor,  a  carbon-steel  turning  tool  is  used. 
The  cutting  speed  is   18  ft.  per  minute.     How  many  revolutions  per 
minute  should  the  work  make  if  the  arbor  is  3  in.  in  diameter? 

Ans.  22.9. 

4.  A  J-in.  drill,  cutting  cast  iron,  may  cut  at  the  rate  of  40  ft.  per  min- 
ute.    How  many  revolutions  per  minute  can  it  make?       Ans.  203.7. 


CIRCLES  171 

6.  In  turning  a  car  wheel  27  in.  in  diameter,  it  makes  If  revolutions 
per  minute.     What  is  the  speed  of  the  cutting  tool? 

Ans.  12.37  ft.  per  minute. 

6.  How  long  will  it  take  to  turn  off  one  layer  from  the  surface  of  a 
car  wheel  4  in.  thick  and  30  in.  in  diameter,  if  the  cutting  is  15  ft.  per 
minute  and  the  feed  |  in.  ?  Ans.  16f  minutes  nearly. 

7.  How  long  would  it  require  to  make  one  cut  over  the  surface  of  a 
tool-steel  arbor  2  in.  in  diameter  and  10  in.  in  length,  if  the  cutting  speed 
is  18  ft.  per  minute  and  the  feed  of  the  cutting  tool  ^  in.  per  revolution 
of  the  work?  Ans.  4.65+  minutes. 

8.  In  Kent's  Mechanical  Engineer's  Pocket-book  are  given  the  follow- 
ing formulas  for  finding  results  in  cutting  speed  problems: 

Let    d  =  the  diameter  of  the  rotating  piece  in  inches, 
n=the  number  of  revolutions  per  minute,  and 
£=the  cutting  speed  in  feet  per  minute,  then 

„     irdn     nna-ioj  &  3.S2S       ,     3.82$ 

S=—  =  0.2618dn;    »= 


Show  that  these  are  true  and  apply  them  to  the  preceding  exercises. 

9.  The  diameter  of  a  piece  of  cast  iron  to  be  turned  is  7  in.     If  the 
lathe  makes  22  revolutions  per  minute,  what  is  the  cutting  speed? 

Ans.  40.3+  ft.  per  minute. 

10.  A  piece  of  brass  4  in.  in  diameter  is  making  80  R.  P.  M.  in  a  lathe. 
What  is  the  cutting  speed?  Ans.  83.8—  ft.  per  minute. 

11.  A  wrought-iron  shaft  2  in.  in  diameter  and  30  in.  long  is  turned 
at  a  cutting  speed  of  25  ft.  per  minute  and  a  feed  of  50  in.     Find  the 
time  for  turning  the  shaft.  Ans.  25.1+  minutes. 

Solution.     30-5-3*5  =  1200  =  number  of  revolutions. 

2X3.1416X1200     0_  1QOC  . 

—  —       =25.  Io28  =number  of  mm. 

1  2i  X  ^o 

12.  The  cutting  speed  in  a  certain  case  must  not  exceed  40  ft.  per 
minute.     The  piece  to  be  turned  is   If  in.   in  diameter.     How  many 
revolutions  per  minute  can  it  make?  Ans.  87.3. 

13.  Give  to  the  nearest  sixteenth  of  an  inch  the  length  of  a  f-in.  steel 
rod  that  is  turned  per  minute,  if  the  cutting  speed  is  36  ft.  per  minute 
and  the  feed  -^  in.  Ans.  7^  in. 

14.  In  turning  a  car  wheel  3  ft.  in  diameter,  the  highest  rate  of  speed 
allowable  for  the  cutter  is  40  ft.  per  minute.     How  many  revolutions 
per  hour  can  the  wheel  make?  Ans.  254.6. 

15.  A  car  axle  may  be  turned  with  the  cutter  moving  9  ft.  per  minute. 
If  the  axle  is  4J  in.  in  diameter,  how  many  revolutions  can  it  make 
per  minute?  Ans.  7.64. 

BELT  PULLEYS  AND  GEAR  WHEELS 

131.  The  relation  of  size  and  speed  of  driving  and  driven 
gear  wheels  are  the  same  as  those  of  belt  pulleys.     In  calcu- 


172  PRACTICAL  MATHEMATICS 

lating  for  gears  we  use  the  diameter  of  the  pitch  circle,  or  the 
number  of  teeth  as  may  be  necessary. 

A  mechanic  should  be  able  to  determine  quickly  and  accu- 
rately the  speed  of  any  shaft  or  machine,  and  to  find  the  size 
of  a  pulley  in  order  that  a  shaft  or  machine  may  run  at  a 
desired  speed.  He  should  master  the  principles  underlying 
the  rules  and  formulas  used  as  well  as  know  how  to  use  them. 
It  is  well  then  for  the  student  to  work  many  problems  on  pulley 
speeds  before  special  formulas  are  taken  up.  This  will  help 
him  to  master  the  principles,  and  will  make  him  independent 
of  the  formulas.  It  will  also  put  him  into  position  to  derive 
the  formulas. 

For  a  complete  discussion  of  questions  connected  with 
belts  and  belting  see  any  mechanical  engineer's  handbook. 

EXERCISES  41 

1.  A  shaft  having  a  pulley  6  in.  in  diameter  makes  840  R.  P.  M.     If 
this  speed  is  to  be  reduced  to  320  revolutions,  what  size  of  pulley  should 
be  used? 

Solution.  If  the  6-in.  pulley  makes  840  R.  P.  M.,  a  point  on  the  belt 
moves  6X3.1416X840  in.  per  minute.  Then  in  order  to  make  320 

6X3.1416X840  . 
R.  P.  M.,  the  pulley  must  be  —     — ox~ —    ~  in.  in  circumference,  and 

6X3.1416X840 
hence     320x3  1416    =1^*  m<  m  diameter. 

2.  The  pulley  on  the  armature  shaft  of  a  dynamo  is  4  in.  in  diameter. 
This  is  to  be  belted  to  a  driving  shaft  which  makes  500  revolutions  per 
minute.     The  speed  of  the  dynamo  must  be  1700  revolutions  per  minute. 
What  must  be  the  size  of  the  pulley  placed  on  the  shaft? 

Ana.  13s  in.  in  diameter. 

3.  A  shaft  has  upon  it  two  pulleys,  each  8  in.  in  diameter.     The  speed 
of  the  shaft  is  400  revolutions  per  minute.     What  must  be  the  size  of 
the  pulleys  of  two  machines  if,  when  belted  to  the  shaft,  one  of  them  has 
a  speed  of  300  revolutions  per  minute  and  the  other  900? 

Ans.  10f  in.  and  35  in.  in  diameter. 

4.  The  pulley  on  the  headstock  of  a  lathe  is  3  in.  in  diameter.     This 
is  belted  to  an  8-in.  pulley  on  a  shaft  that  makes  420  revolutions  per  min- 
ute.    At  what  rate  will  a  block  of  wood  placed  in  the  chuck  revolve? 

Ans.  1120  R.  P.  M. 

6.  If  the  wheels  of  an  electric  car  are  2  ft.,  the  axle  cogwheel  8  in.,  and 
the  cogwheel  attached  to  the  motor  12  in.  in  diameter,  what  must  be 
the  Hpeed  of  the  motor  to  carry  the  car  a  mile  in  5  minutes? 

Atu.  112.04+  R.  P.  M. 


CIRCLES 


173 


6.  In  two  connected  belt  pulleys,  or  gear  wheels,  if  D  is  the  diameter 
of  the  driving  wheel,  d  the  diameter  of  the  driven  wheel,  R  the  number  of 
revolutions  per  minute  of  driver,  and  r  the  number  of  revolutions  per 
minute  of  driven,  find  r  in  terms  of  D,  d,  and  R.  .  ,  _^5. 

d 

Discussion.     In  Fig.  113,  A  is  the  driving  pulley  and  B  is  the  driven 
pulley.     It  is  evident  that,  since  the  belt 
does  not  slip,  a  point  on  the  circumfer- 
ence of  B  must  move  as  far  in  a  minute 
as  a  point  on  the  circumference  of  A. 

Since  A  makes  R  revolutions  per  min- 
ute, a  point  on  its  circumference  will 
move  RirD  units  per  minute.  Similarly 
a  point  on  the  circumference  of  B  will 
move  rird  units  per  minute. 


FIG.   113. 


i       i       RD 
or  RD  =  rd  and  r  =  —,—• 

d 

7.  In  any  system  of  pulleys  or  gears,  the  general  rule  holds:  that  the 
product  of  the  diameters,  or  numbers  of  teeth,  of  the  driving  wheels  and 
the  number  of  revolutions  per  minute  of  the  first  driver  must  be  equal 
to  the  product  of  the  diameters,  or  the  numbers  of  teeth,  of  the  driven 
wheels  and  the  number  of  revolutions  per  minute  of  the  last  driven  wheel. 
As  a  formula  this  may  be  stated 

=  RXDXD'XD"XD'"Xetc. 

dXd'Xd"Xd'"Xetc. 

where  D,  D',  D",  etc.,  are  the  diameters  of  the  driving  pulleys,  d,  d'  d", 
etc.,  are  the  diameters  of  the  driven  pulleys,  R  is  the  R.  P.  M.  of  the  first 
driver,  and  r  is  the  R.  P.  M.  of  the  last  driven  pulley.  Show  why  this 
is  true. 

8.  The  number  of  revolu- 
tions the  governor  of  a  steam 
engine  is  intended  to  run  is 
given  by  the  builder.  If  the 
speed  of  the  governor  is  120 
R.  P.  M.,  size  of  governor 
pulley  8  in.,  and  the  desired 
speed  of  the  engine  90  R.  P.  M., 
find  the  diameter  of  the  pulley  to  be  put  on  the  engine  shaft  to  run  the 
governor  pulley.  Ans.  lOf  in. 

9.  An  endless  knife  runs  on  pulleys  48  in.  in  diameter  as  shown  in 
Fig.  114,  at  a  rate  of  180  R.  P.  M.  If  the  pulleys  are  decreased  18  in. 
in  diameter,  how  many  R.  P.  M.  will  they  have  to  make  to  keep  the 
knife  traveling  at  the  original  speed? 


Knife 


174 


PRACTICAL  MA  Til  EM  A  TICK 


Solution.     180X3.1416X48  in.  =  rate  per  minute. 
3.141GX30  in.  =  circumference  of  reduced  pulleys. 

180X3.1410X48 

'FTTo —  =288=  number  of  R.P.M.  of  reduced  pulleys. 


FIG.  115. 


FIG.  11G. 


Engine  Shaft   F 


UjD" 


Line 

Shaft 


Fio.  117. 

10.  Adapt  the  formula  of  exercise  7  to  the  following :  A  train  of  wheels 
consists  of  four  wheels  each  12  in.  in  diameter  of  pitch  circle,  and  three 


CIRCLES  175 

pinions  4  in.,  4  in.,  and  3  in.  in  diameter  respectively.  The  first  three 
large  wheels  are  the  drivers  and  the  first  makes  36  revolutions  per  minute. 
Required  the  speed  of  the  last  wheel.  Ans.  1296  R.  P.  M. 

11.  In  the  train  of  the  preceding  exercise,  what  is  the  speed  of  the  first 
large  wheel  if  the  pinions  are  the  drivers,  the  3-in.  pinion  being  the  first 
driver  and  making  36  revolutions  per  minute?  Ans.  1  R.  P.  M. 

12.  Pulleys  are  arranged  as  in  Fig.  115.     Pulley  A  makes  192  R.  P.  M., 
is  the  driver,  and  is  14  in.  in  diameter.     Pulley  B  is  8  in.  in  diameter. 
Pulley  C  is  6  in.  in  diameter  and  is  to  make  a  required  1400  R.  P.  M. 
Find  the  diameter  to  make  the  pulley  D,  fastened  to  the  same  shaft  as 
B,  in  order  that  C  may  have  the  desired  number  of  revolutions  per  minute. 

Ans.  25  in. 

13.  Find  the  number  of  R.  P.  M.  of  the  last  gear  shown  in  Fig.  116, 
if  the  gear  having  84  teeth  makes  36  R.  P.  M. 

14.  In  Fig.  117,  if  a  160-in.  pulley  on  the  engine  shaft  drives  a  60-in. 
pulley  on  the  line  shaft,  and  a  40-in.  pulley  on  the  line  shaft  drives  an 
18-in.  pulley  on  the  counter  shaft,  find  the  number  of  revolutions  per 
minute  of  the  counter  shaft  if  the  engine  shaft  runs  at  80  R.  P.  M. 

Ans.  474  nearly. 

THE  MIL 

132.  The  circular  mil. — In  most  cases,  electrical  conductors 
have  a  circular  cross  section.  We  know  that  the  area  of  a 
circle  is  found  by  the  formula  A  =  j-rrd2,  which  brings  in  the 
inconvenient  factor  fr,  or  0.7854.  In  order  to  avoid 
this  factor,  a  new  unit  has  been  adopted  for  commercial  work. 
This  unit  is  the  circular  mil  (abbreviation  C.  M.)  which  is 
the  area  of  a  circle  one  mil,  or  0.001  inch,  in  diameter. 

If  A  is  the  area  in  circular  mils  of  any  circle  and  d  the  di- 
ameter in  mils,  then,  since  a  circle  1  mil  in  diameter  has  an  area 
of  1  C.  M.,  we  have  the  proportion 

1  =T 
Ad2 

for  the  areas  of  the  circles  are  in  the  same  ratio  as  the  squares 
of  the  diameters.  This  proportion  gives 

A  =d2. 
This  stated  in  words  is  the  following: 

RULE.  The  area  of  a  circle  in  circular  mils  is  the  square  of 
the  diameter  in  mils,  or  thousandths  of  an  inch. 

Thus,  a  0000  gage  B.  and  S.  wire  is  0.46  in.  =460  mils  in  diameter, 
and  hence  has  an  area  of  4602  =211,600  C.  M. 


176  PRACTICAL  MATHEMATICS 

If  the  area  in  circular  mils  is  given,  the  diameter  in  mils  can 
evidently  be  found  by  taking  the  square  root  of  the  area,  or 


133.  The  square  mil.  —  The  square  mil  is  sometimes  used, 
and  is  the  area  of  a  square  1  mil  on  a  side.  Since  the  area  of 
a  circle  is  0.7854d2,  it  is  seen  that  0.7854  square  mil  =  1  C.  M. 

EXERCISES  42 

1.  Find  the  number  of  circular  mils  in  the  area  of  B.  and  S.  gage  wires 
Nos.  40,  20  and  10.     (See  Table  VII.) 

Ans.  9.88  +  J  1021.5  +  ;  10,383-. 

2.  How  many  square  mils  in  a  bar  J  in.  by  f  in.  in  cross  section? 

Ans.  187,500. 

3.  How  many  circular  mils  are  equal  to  20,000  square  mils? 

Ans.  25,464.8-. 

4.  Find  the  diameter  in  mils  and  in  inches  of  a  circular  rod  having  a 
cross  section  of  237,600  C.  M.  Ans.  487.4+  mils;  0.4874+  in. 

5.  In  ordinary  practice,  trolley  wire  is  0  or  00  B.  and  S.  hard-drawn 
copper  wire.     What  is  the  area  of  the  cross  section  of  each  in  circular 
mils?  Ans.  105,535-   C.  M.  or  133,076+  C.  M. 


CHAPTER  XIII 
GRAPHICAL  METHODS 

ANGLES 

134.  Units. — In  measuring  any  magnitude,  a  unit  of  measure 
is  necessary.     In  measuring  length,  there  are  various  units, 
as  the  inch,  foot,  meter,  and  mile.     Likewise  in  the  measure- 
ment of  angles,  there  are  in  use,  as  units,  the  right  angle,  the 
degree,  and  the  radian. 

Right  angle  as  unit.  When  using  the  right  angle  as  a  unit, 
we  speak  of  an  angle  as  such  a  part  of,  or  as  so  many  times,  a 
right  angle. 

The  degree  as  a  unit.  The  degree  as  a  unit  for  measuring 
angles  may  be  defined  as  the  value  of  the  angle  formed  by 
dividing  a  right  angle  into  90  equal  parts.  The  degree  is  also 
used  as  a  unit  for  measuring  arcs.  It  is  then  defined  as  -$1-$ 
part  of  a  circumference.  In  either  case  the  degree  is  divided 
into  60  parts  called  minutes,  and  the  minute  into  60  parts 
called  seconds.  Degrees,  minutes,  and  seconds  of  angle  or 
arc  are  indicated  by  the  signs  °,  ',  and  ". 

For  example,  a  measurement  of  27  degrees,  47  minutes,  35  seconds  is 
written  thus:  27°  47'  35". 

As  already  defined,  if  an  angle  has  its  vertex  at  the  center 
of  a  circle  and  its  sides  formed  by  the  radii  of  the  circle,  it  is 
spoken  of  as  an  angle  at  the  center  of  a  circle.  The  number  of 
degrees  in  the  angle  so  placed  is  equal  to  the  number  of  degrees 
in  the  arc  of  the  circle  intercepted  between  the  sides  of  the 
angle. 

Thus,  in  Fig.  118,  AOB  is  an  angle  at  the  center.  The  number  of 
degrees  in  this  angle  equals  the  number  of  degrees  in  the  arc  AB. 

The  angle  AOB  is  said  to  be  measured  by  the  arc  AB. 

135.  Circular  measure,  radian. — The  unit  of  circular  meas- 
ure of  angles  is  the  radian.     The  radian  is  defined  as  the 

177 


178 


PRACTICAL  MA THEMATICS 


angle  which  at  the  center  of  a  circle  is  measured  by  an  arc 
equal  in  length  to  the  radius  of  the  circle. 

In  Fig.  US,  arc  ,4/?  =  rndius  OA,  hence  angle  AOB  is  one  radian. 

Since  a  circumference  is  2r  times 
the  radius,  there  are  2*  arc  lengths 
equal  to  the  radius  in  a  circumfer- 
ence; and  hence  2w  radians  are  meas- 
ured by  the  circumference  of  a  circle, 
or  2ir  radians  =  360°. 

From  this,  T  radians  =  360°  -f-  2  = 
180°,  and 

1  radian  =  180°  -5-  TT  =  57.29578°  - . 
Reduced  to  degrees,  minutes,  and  seconds: 

1  radian  =  57°  17'  44.8". 
In  a  similar  manner,  if  180°  =T  radians,  then 

I°=TT  radians -5-180  =  0.01745+  radians. 
136.  The  protractor. — The  protractor  consists  of  a  chcular 
or  semicircular  scale  of  convenient  diameter.     The  circum- 
ference of  this  scale  is  divided  into  degrees,  half  degrees,  and 
sometimes  into  quarter  degrees,  as  shown  in  Fig.  119.     The 


FII;.  118 


divisions  of  the  scale  are  numbered  from  0°  to  180°,  beginning 
at  each  end.  The  sum  of  the  two  readings  at  any  point  is 
180°.  This  method  of  numbering  enables  one  to  measure  an 
angle  from  either  end  of  the  protractor,  or  it  enables  one  to 
set  off  angles  in  either  a  right-  or  a  left-hand  direction. 


GRAPHICAL  METHODS  179 

137.  To   measure   an   angle   with   protractor. — Place   the 
protractor  over  the  angle  to  be  measured,  so  that  the  point 
C  is  on  the  vertex  of  the  angle,  and  either  half  of  the  side  A B 
will  fall  upon  one  side  of  the  angle.     The  other  side  of  the 
angle  should  pass  through  some  mark  on  the  scale  of  the  pro- 
tractor.    The  reading  on  the  scale  where  this  side  of  the 
angle  crosses  it  is  the  measure  of  the  angle  in  degrees.     In  work 
with  the  protractor,  a  hard  pencil  with  sharp  point  should  be 
used. 

138.  To  lay  off  an  angle  with  a  protractor. — Draw  one  side 
of  the  angle  and  locate  the  vertex.     Place  side  AB  of  protractor 
on  side  drawn  and  point  C  on  the  vertex.     Locate  reading  of 
value  of  angle  required  on  scale  of  protractor,  and  connect  this 
with  the  vertex.     The  degree  of  accuracy  with  which  an  angle 
can  be  laid  off  depends  upon  the  instruments  and  the  one  who 
uses  them,  but  largely  upon  the  size  of  the  protractor.     It  is 
well  then  to  use  a  fairly  large  protractor,  one  5  or  6  in.  in 
diameter. 

EXERCISES  43 

1.  Draw  three  triangles  and  measure  their  angles.     Find  the  sum  of 
the  angles  of  each  triangle.     Are  their  sums  equal? 

2.  Draw  a  right  triangle  and  measure  the  two  acute  angles.     What  is 
their  sum? 

3.  Draw  an  equilateral  triangle  and  measure  the  angles.     What  is 
their  sum? 

4.  Draw  an  isosceles  triangle  and  measure  its  angles.     Are  there  two 
equal  angles? 

6.  Construct  angles  of  26°,  75°  30',  106°  45',  and  146°  15'. 

6.  Describe  how  an  angle  of  If  radians  may  be  constructed. 

7.  How  many  radians  in  each  of  the  angles:  27°  45',  47°  26',  109°  30'? 

Arcs.  0.484  +  ;  0.8279  -;  1.911  +  . 

8.  What  is  the  value  in  degrees  and  decimals  of  degrees  of  each  of  the 
angles:  2  radians,  f  radian,  1.75  radians? 

Ans.  114.5916°-;    42.9718°  +  ;    100.2676° -i-. 

9.  Draw  a  line  AB  2|  in.  long;  at  A  lay  off  an  angle  of  115°  30'  and 
draw  line  AC  f  in.  long;  at  B  lay  off  an  angle  of  97°  and  draw  line  BD 
If  in.  long.     Find  length  of  CD  by  measuring. 

10.  Draw  a  parallelogram  and  measure  its  angles.     \Vhat  is  their  sum  ? 
How  do  the  angles  compare? 

11.  Draw  a>  quadrilateral   and   measure  its  angles.     What  is  their 
sum? 


1 80  1'UA  CTJCAL  MA  THEM  A  TICS 

ANGLES  BY  CHORDS 

139.  An  angle  can  usually  be  laid  off  more  accurately  by 
means  of  a  table  of  chords  than  by  the  protractor,  and  the 
method  is  often  more  convenient.     While  with  the  protractor 
it  is  not  easy  to  measure  more  accurately  than  to  half  degrees, 
with  the  table  of  chords  the  angle  may  be  laid  off  to  tenths  of 
a  degree. 

In  Table  VI  are  given  the  lengths  of  the  chords  in  a  circle  of 
radius  1,  for  central  angles  from  0°  to  89.9°. 

140.  To  find  a  chord  length  from  the  table. — To  find  the 
length  of  the  chord  for  an  angle  of,  say,  27.6°,  find  20°  in 
the  left-hand  column  of  the  table;  go  to  the  right  to  column 
headed  7°  and  there  find  0.467,  the  length  of  the  chord  for  27°. 
Now  add  to  0.467  the  number  0.010,  found  still  further  to  the 
right  in  the  column  headed  0.6°.     This  gives  0.477  as  the 
length  of  the  chord  for  27.6°.     This  means  that,  if  the  vertex 
of  the  angle  is  at  the  center  of  a  circle  of  radius  1  in.,  the  chord 
has  a  length  of  0.477  in. 

For  the  length  of  the  chord  in  a  circle  of  any  other  radius, 
multiply  this  chord  by  the  length  of  the  radius  of  the  given 

circle.  Thus,  for  a  circle  of  6-in. 
radius,  the  chord  for  a  central  angle 
of  27.6°  is  6X0.477  in.  =  2.862  in. 

141.  To  lay  off  an  angle. — How 
this  can  be  done  is  best  shown  by 
an  actual  construction. 

Example.     Lay  off  an  angle  of  48.3°. 
From  Table  VI,  the  chord  for  48.3° 
is  0.818.     Draw  a  line  OP,  Fig.  120. 
i-'»>.  With  0  as  a  center  and  radius  1  in., 

draw  the  arc  AB.     With  A  as  a  center 

and  with  radius  0.818  in.,  strike  the  arc  cutting  AB  at  C. 
Connect  0  with  C,  and  AOC  is  the  angle  48.3°  required. 

For  this  construction  there  are  required  a  pair  of  compasses, 
a  ruler  divided  into  0.01  in.,  and  a  sharp  pencil. 

An  angle  larger  than  a  right  angle  may  be'  laid  off  by  first 
constructing  a  right  angle,  and  then  laying  off  the  remainder 
adjacent  to  this. 


GRAPHICAL  METHODS  181 

The  table  can  be  conveniently  used  if  a  radius  is  taken  10 
units  in  length. 

142.  To  measure  an  angle. — With  the  vertex  of  the  angle 
as  center  and  radius  1,  draw  an  arc  cutting  both  sides  of  the 
angle.     Measure  the  chord  drawn  between  these  two  points, 
and  refer  to  the  table  of  chords  to  find  the  angle. 

An  angle  larger  than  90°  may  be  measured  by  first  laying 
off  an  angle  of  90°,  and  then  measuring  the  remaining  angle. 

EXERCISES  44 

1.  In  a  circle  of  radius  1  in.,  find  the  lengths  of  the  chords  for  the 
following  angles: 

(a)  43.8°.  Ans.  0.746  in.  (e)  63°  48'.  Ans.  1.057  in. 

(b)  29.1°.  Ans.  0.503  in.  (f)   10°  40'.  Ans.  0.186  in. 

(c)  13.3°.  Ans.  0.231  in.  (g)  43°  30'.  Ans.  0.741  in. 

(d)  79.2°.  Ans.   1.275  in.  (h)  78°  15'.  Ans.  1.263  in. 

2.  Construct   the   following   angles:    (a)    60°,    (b)   10°  45',  (c)  72.8°, 
(d)  115°  30',  (e)  145.3°,  (f )  39.9°. 

3.  Draw  the  chords  in  a  circle  of  1  in.  radius  to  form  a  regular  polygon 
of  5  sides.     (Each  angle  at  the  center  is  72°.) 

4.  Inscribe  regular  polygons  of  6,  7,  8,  9,  10,  and  15  sides  in  a  circle  of 
radius  3  in. 

Suggestion.  Divide  360°  by  the  number  of  sides  to  find  the  angle 
at  the  center  of  the  circle  determined  by  the  side.  From  Table  VI 
determine  the  length  of  the  side,  and  space  off  the  circumference  using 
this  length. 

If  the  number  of  sides  is  9,  the  angle  at  center  is  360°  s-9  =  40°.  When 
the  radius  of  the  circle  is  3,  the  side  of  the  polygon  is  2.052. 

AREAS,  GRAPHICAL  METHODS 

143.  Drawing  to  scale. — If  we  wish  to  draw  to  scale  the 
floor  of  a  room  10  ft.  by  20  ft.,  we  may  conveniently  represent 
1  ft.  in  the  dimensions  of  the  room  by  |  in.  in  the  drawing. 
Whatever  dimension  is  measured  in  the  drawing  and  given 
in  eighths  of  an  inch  can  be  interpreted  as  feet  when  applied 
to  the  floor.     The  map  of  a  country  may  be  drawn  on  a  scale 
of  50  miles  to  an  inch,  or  any  other  convenient  scale. 

144.  To  construct  a  triangle  having  given  two  sides  and 
the  angle  between  these  sides. — Let  the  side  AB  =  10  ft., 
and  AC  =  8  ft.,  and  the  angle  A  between  these  sides  be  47°  45'. 


182 


PRACTICAL  MATHEMATICS 


Choose  a  convenient  scale,  say,  J  in.  for  1  ft.  Draw  line 
AB,  Fig.  121,  in  length  Y  in.;  lay  off  angle  BAC=47°  45'; 
make  AC  |  in.;  and  draw  CB.  Then  the  triangle  ACB  is 
the  required  representation  of  the  triangle. 

If  it  is  required  to  find  the  area  of  the  triangle,  the  measured 


length  of  the  altitude  DC  in  eighths  of  an  inch  times  one-half 
of  the  base  AB  in  the  same  unit,  would  give  the  area  of  the 
triangle  in  square  feet. 

145.  To  construct  a  triangle  when  given  two  angles  and 
the  side  between  these  angles. — Let  the  side  AB  =  30  ft.,  the 
angle  A  =  37°,  and  the  angle  B  =  49°.  Choose  a  scale,  say, 
iV  in.  for  1  ft.  Draw  line  AB,  Fig.  122,  in  length  f  g  in.;  lay 
off  angle  A  =37°,  and  #  =  49°;  and  extend  the  sides  of  these 

angles  till  they  meet.  Then 
the  triangle  A  CB  is  the  required 
triangle  represented  on  a  scale 
of  TV  in.  to  1  ft.  The  area  of 
this  triangle  can  easily  be  found 
by  measuring  the  altitude  DC 
and  applying  the  rule  for  the 
area  of  a  triangle. 

146.  To  construct  a  triangle 
when  the  three  sides  are 
given. — Let  AB  =  6Q  rods, 
AC  =  70  rods,  and  BC  =  80 
rods.  Choose  a  scale  of,  say, 
40  rods  to  the  inch.  Then  60  rods  is  represented  by  1^  in., 
70  rods  by  If  in.,  and  80  rods  by  2  in.  Draw  AB,  Fig.  123, 
in  length  l£  in.  With  the  compasses  and  a  radius  of  If  in. 
draw  an  arc  with  A  as  center.  With  B  as  center,  and  radius 


FIG.   123. 


183 


2  in.,  draw  an  arc  to  intersect  this  at  C.  Draw  the  lines 
AC  and  BC.  Then  triangle  ABC  is  the  required  triangle. 
In  a  similar  manner  other  shaped  figures  may  be  constructed 
to  scale.  In  many  cases  these  drawings  may  be  divided  into 
triangles,  squares,  and  rectangles,  which  may  be  measured, 
and  so  the  entire  area  of  the  figure  be  found. 

D  n 


FIG.    124. 

Example.  A  piece  of  ground  in  the  form  of  a  quadri- 
lateral is  represented  by  Fig.  124  to  a  scale  of  40  rods  to 
an  inch.  The  area  can  be  found  by  drawing  the  diagonal 
AC  and  the  altitudes  of  the  triangles  A  CD  and  ABC.  The 
sum  of  the  areas  of  these  tri- 
angles equals  the  area  of  the 
quadrilateral  ABGD. 

147.  Areas  found  by  the  use 
of  squared  paper. — It  is  often 
convenient  to  find  the  area  of 
an  irregular  figure  by  drawing  it 
on  squared  paper  (paper  accu- 
rately ruled  into  small  squares). 
The  figure  will  usually  be  drawn 
to  some  scale  that  uses  the  side 
of  one  of  the  small  squares  as 
a  unit. 

The  method  is  most  nearly  accurate  on  irregular  figures, 
and  is  liable  to  considerable  error  when  the  boundary  has 
long  straight  lines,  nearly  parallel  to  the  lines  forming  the 
squares. 


FIG.   125. 


1M  PRACTICAL  MATHEMATICS 

As  an  illustration  of  the  method,  find  the  area  of  the  circle 
in  Fig.  125,  if  the  circle  is  drawn  on  a  scale  of  1  in.  to  a  side 
of  the  squares. 

First,  determine  how  many  squares  are  wholly  within  the 
circle. 

Second,  count  as  whole  squares  the  squares  that  are  half  or 
more  than  half  within  the  circle,  and  neglect  those  squares  that 
are  less  than  half  within  the  circle. 

Here  it  is  most  convenient  to  count  the  squares  in  one  quar- 
ter of  the  circle,  and  then  multiply  by  4  to  get  the  area  of  the 
whole  circle. 

From  the  figure,  it  is  seen  that  there  are  30  whole  squares  in 
the  quarter  of  the  circle  AOB.  Counting  the  squares  marked 
1,  gives  8  partial  squares  to  be  taken  as  whole  squares.  The 
squares  marked  0  are  not  counted. 

30+8  =  38  squares  for  the  quarter  circle. 
38X4  =  152  squares  for  the  circle. 
.'.  area  of  the  circle  is  152  in.2 
By  formula  [24],  area  =  3.1416X72=154-  in.2 

148.  Other  methods  for  approximating  areas. — (a)  The 
planimeter  is  an  instrument  for  estimating  areas.  There 
are  several  forms  of  this  device;  but,  as  instructions  for  its 
use  are  given  with  each  instrument,  it  will  not  be  described 
here. 

(6)  The  area,  when  very  irregular,  can  often  be  estimated 
quite  accurately  by  cutting  full  size  or  to  scale  out  of  card- 
board or  sheet  tin.  Weigh  on  accurate  scales  the  piece  of 
tin  or  cardboard ;  also  weigh  a  square  unit  of  the  same  material. 
Divide  the  weight  of  the  piece  by  the  weight  of  the  square 
unit.  The  quotient  is  the  number  of  square  units  in  the 
figure. 

(c)  Other  methods  will  be  found  in  Chapt.  XXXIII. 

EXERCISES  46 

1.  The  two  sides  of  a  triangle  are  18  ft.  and  24  ft.,  respectively,  and 
include  an  angle  of  98°.  Find  the  length  of  the  other  side,  and  find  the 
area  of  the  triangle  by  two  separate  sets  of  measurements,  that  is,  draw 
two  altitudes  and  use  two  sides  as  bases. 


GRAPHICAL  METHODS 


185 


2.  The  base  of  a  triangle  is  10  ft.,  and  the  other  two  sides  are  7  ft.  and 
5  ft.  respectively.     Find  graphically  the  length  of  the  altitude  to  the 
base.     What  is  the  area  of  the  triangle? 

3.  Find  the  area  of  the  triangle  in  exercise  2  by  rule  for  the  area  of  a 
triangle  when  the  three  sides  are  given.  Ans.  16.25—  ft.2 

4.  In  a  triangle,  one  side  is  28  ft.  and  the  adjacent  angles  are  39°  45' 
and  49°  30'.     Find  the  lengths  of  the  other  sides  and  the  area  of  the 
triangle. 

5.  The  angles  of  a  triangle  are  48°,  78°,  and  54°.     Find  the  length  of 
the  side  opposite  the  angle  78°,  if  the  side  opposite  48°  is  32  ft. 

6.  The  two  sides  AB  and  BC  of  a  triangle 
are  44.7  ft.  and  96.8  ft.,  respectively,  the  angle 
ABC  being  32°.     Find    (a)   the  length  of  the 
perpendicular  drawn  from   A  to  BC;   (b)  the 
area  of  the  triangle  ABC]  (c)  the  angles  at  A 
and  C. 

Ans.  (a)  23.69  ft.  ;<b)  1147ft.2;  (c)  22°;  126°. 

7.  Find  the  area  of  AbcD,  Fig.  126,  which  is 

on  a  scale  of  1  in.  to  the  side  of  a  square,  by  counting  the  squares. 

8.  Find  the  area  of  the  ellipse  in  Fig.   127,  which  is  on  a  scale  of 
4  rods  to  the  side  of  a  square,  by  counting  squares.     Find  the  area  by 
formula  [31],  and  compare  results. 

9.  The  quadrilateral  in  Fig.  128  is  on  a  scale  of  16  rods  to  the  inch. 
Find  its  area  by  dividing  into  triangles. 

10.  Draw  a  triangle  on  a  scale  of  j  in.  to  the  foot,  having  sides  of 
17  ft.,   19  ft.,  and  23  ft.  respectively.     Draw  any  altitude  of  the  tri- 
angle,  measure  it,   and  compute  the  area  of  the  triangle.     Check  by 
drawing  the  other  altitudes  and  computing  the  area  again. 


FIG.   126. 


FIG.  127. 


FIG.   128. 


11.  Draw  the  following  to  scale,  using  ^  in.  to  the  rod,  then  find 
the  area  in  acres.     Start  at  a  point  A,  go  east  20  rd.  to  B,  north  10  rd. 
to  C,  east  10  rd.  to  D,  north  40  rd.  to  E,  west  40  rd.  to  F,  south  20  rd. 
to  G,  east  20  rd.  to  H,  and  then  to  A.  Ans.  9^  A. 

12.  Find  the  area  of  the  following  plot  of  ground:  Start  at  A,  go  east 
10  ft.  to  B,  north  20  ft.  to  C,  northeast  14.14  ft.  to  D,  north  10  ft.  to  E, 
west  20  ft.  to  F,  and  south  to  A.  Ans.  550  ft.2 


186 


PRACTICAL  MATHEMATICS 


VARIOUS  USEFUL  CONSTRUCTIONS 

149.  To  divide  a  line  of  any  length  into  a  given  number  of 
equal  parts. — In  Fig.  129,  let  AB  be  the  line  which  it  is  re- 
quired to  divide  into  seven  equal  parts.  Draw  a  line  AC, 
making  any  convenient  angle  with  AB.  Take  some  con- 
venient length,  as  a  half  inch,  and  beginning  at  A  mark  seven 
of  these  lengths  on  AC.  This  determines  the  points  a,  b,  c, 


FIG.  129. 

d,  e,  f,  and  g.  Draw  a  line  from  g  to  B,  and  draw  lines 
parallel  to  gB  through  /,  d,  c,  b,  and  a.  These  lines  divide 
AB  into  seven  equal  parts. 

It  is  readily  seen  that  this  method  can  be  used  to  divide 
any  line  into  any  number  of  equal  parts. 


Fio.  130. 

Example.  The  carpenter  makes  use  of  this  construction 
when  he  uses  his  steel  square,  as  shown  in  Fig.  130,  to  divide 
a  board  of  any  width  into  any  number  of  equal  strips.  Here 
un  8-in.  board  is  to  be  divided  into  five  equal  strips.  A  number 
of  inches  divisible  by  5,  as  10  in.,  is  taken  on  the  square,  and  the 
square  is  placed  as  shown.  A  mark  is  made  on  the  board  at 


GRAPHICAL  METHODS 


187 


FIG.  131. 


the  2-in.  divisions  on  the  square.     This  divides  the  board 
as  desired. 

150.  To  cut  off  the  corners  of  a  square  so  as  to  form  a 
regular  octagon.  —  Draw  the  diagonals  of  the  square,  Fig.  131. 
With  the  compasses  open  a  distance 

AO,  one-half  a  diagonal,  and  with 
each  vertex  of  the  square  as  a  center, 
strike  the  arcs  at  a  and  a',  b  and  V, 
c  and  c',  d  and  df.  Connect  these 
points  as  shown  in  the  figure,  and  get 
the  regular  octagon  bac'b'dca'd'. 

151.  To  divide  a  given  circle  into 
any  number  of  equal  parts  by  con- 
centric circles.  —  Let  the  largest  circle 

of  Fig.  132  be  the  given  circle,  and  let  it  be  desired  to  divide 

it  into  four  equal  parts.     Draw  the  radius  OA  and  divide  it 

into  the  same   number   of  equal  parts.     Draw  a  semicircle 

on  this  radius  as  a  diameter. 

Erect  perpendiculars  to  OA  at  each  division  point  a,  b, 

and  c,  and  let  them  inter- 
sect the  circumference  of 
the  semicircle  at  a',  bf, 
and  cf  respectively.  Using 
Ocf,  Ob',  and  Oaf  as  radii 
and  0  as  a  center,  draw 
the  three  concentric  circles. 
These  circles  divide  the 
original  circle  as  desired. 

152.  To  inscribe  regu- 
lar polygons.  —  (See  Arts. 
127  and  140.)  (1)  To  lay 
out  a  square  in  a  circle. 

Draw  two  perpendicular   diameters    as  shown  in  Fig.    133, 

and  connect  their  successive  extremities.     This  gives  the  in- 

scribed square  ACBD. 

(2)   To  lay  out  a  pentagon  in  a  circle.     Draw  two  perpen- 

dicular diameters  AB  and  CD,  Fig.   134,  bisect  AO  at  F. 

With  E  as  a  center  and  ED  as  a  radius,  draw  the  arc  DF. 

The  length  DF  is  equal  to  the  side  of  the  inscribed  pentagon. 


FlG-  132- 


188 


PRACTICAL  MATHEMATICS 


(3)  To  lay  out  a  hexagon  in  a  circle.     The  radius  of  the 
circle  is  equal  to  a  side  of  the  inscribed  hexagon. 

(4)  To  lay  out  an  equilateral  triangle  in  a  circle.     Connect 
the  alternate  vertices  of  the  hexagon  as  shown  in  Fig.  135. 

(5)  To  lay  out  a  regular  heptagon  in  a  circle.     Make  a  con- 


struction as  shown  in  Fig.  136,  and  AB  is  very  nearly  the 
side  of  the  inscribed  regular  heptagon. 

(6)  To  lay  out  a  regular  octagon  in  a  circle.     Bisect  the 
arcs  of  the  inscribed  square. 

(7)  To  lay  out  a  regular  decagon  in  a  circle.     Bisect  the 


Fio.   135. 


Fio.  136. 


arcs  of  the  inscribed  regular  pentagon. 

153.  To  draw  the  arc  of  a  segment  when  the  chord  and 
the  height  of  the  segment  are  given. — This  method  is  to 
be  used  when  it  is  not  convenient  to  find  the  radius  and  use 
it.  In  Fig.  137,  AB  is  the  chord  and  FC  is  the  height.  Draw 


GRAPHICAL  METHODS 


189 


AC',  AD  perpendicular  to  AC]  CD  parallel  to  FA;  AE  per- 
pendicular to  AF]  and  divide  AF,  DC,  and  AE  into  the  same 
number  of  equal  parts.  Letter  them  as  in  the  figure.  Draw 
da,  eb,  fc,  Cl,  C2,  and  C3.  The  points  of  intersection  of 
these  are  points  on  the  arc.  The  more  equal  parts  the  lines 
are  divided  into,  the  more  points  of  the  arc  will  be  determined. 


164.  To  find  the  radius  of  a  circle  when  only  a  part  of 
the  circumference  is  known. — It  often  happens  that  one  has 
a  part  only  of  a  wheel  or  pulley  from  which  he  must  determine 
the  size  of  a  new  wheel  or  pulley.  The  method  by  bisecting 
the  arcs  is  shown  in  Fig.  138.  Let  ABC  be  the  arc  given. 
Draw  two  chords  AB  and  BC  of  any  convenient  lengths. 
Draw  perpendicular  bisectors  of  these.  They  will  intersect 


at  a  point  0  which  is  the  center  of  the  circle  of  which  the  given 
arc  is  a  part. 

What  amounts  to  the  same  thing  and  is  more  quickly 
done  is  shown  in  Fig.  139.  Draw  three  equal  intersecting 
circles  with  their  centers  on  the  arc,  then  the  lines  drawn 
through  the  intersecting  points  as  shown  meet  at  the  center 
of  the  circle  of  which  the  given  arc  is  a  part. 


190 


PRACTICAL  MATHEMATICS 


155.  How  to  cut  a  strikeboard  to  a  circular  arc. — The 
standard  specifications  of  the  American  Concrete  Institute 
recommend  that  the  surface  of  concrete  highways  be  finished 
to  a  crown  corresponding  to  the  arc  of  a  circle.  The  height 
of  the  crown  at  the  center  of  the  highway  above  the  margin 
of  the  pavement  is  recommended  to  be  not  more  than  -fa 
and  not  less  than  Ti^  of  the  width  of  the  pavement.  A  car- 
penter or  mechanic  can  readily  lay  out  a  curve  of  this  kind 
on  a  strikeboard.  The  method  shown  in  Fig.  140  is  based 
upon  a  simple  geometric  fact  and  if  properly  followed  will 
give  a  true  curve. 

If  the  lower  edge  of  the  piece  from  which  the  strikeboard 
is  to  be  cut  is  not  straight,  the  lay-out  can  be  made  to  a  chalk 


FIG.  140. 

line,  A  B  of  the  figure.  Drive  a  stiff  nail  at  each  of  the  points 
A  and  J3,  which  mark  the  width  of  the  pavement.  Find  the 
mid-point  between  these  nails  and  mark  on  the  board  a  line 
perpendicular  to  the  line  AB.  Drive  a  nail  at  M  as  far  above 
the  mid-point  of  the  line  AB  as  required  by  the  cross  section 
for  the  crown  height  at  center. 

Take  two  light  pieces  of  lumber  which  have  straight  edges, 
and  fasten  them  rigidly  together  so  that  they  form  the  angle- 
frame  A  MK.  The  line  MK  must  be  parallel  to  A  B.  Then  if 
one  side  of  this  angle-frame  is  held  against  the  frame  at  M 
and  one  at  A,  the  intersection  of  the  sides  of  the  angle  will  be 
a  point  of  the  curve  desired.  By  shifting  the  angle-frame,  the 
intersection  may  be  moved  from  M  to  A  and  points  on  the 
curve  marked  as  often  as  desired.  By  turning  the  angle-frame 
over,  the  curve  on  the  other  half  may  be  similarly  marked. 


GRAPHICAL  METHODS  191 

156.  The  vernier. — The  vernier  is  a  device,  invented 
by  Pierre  Vernier,  by  which  measurements  can  be  read  with 
a  much  greater  degree  of  accuracy  than  is  possible  by  mere 
mechanical  division  and  subdivision.  There  are  two  kinds 
of  verniers,  known  as  the  direct  and  reverse.  Only  the  direct 
will  be  described  here. 

The  principle  is  shown  in  its  essentials  in  Fig.  141,  which 
is  a  portion  of  a  graduated  scale,  having  below  it  a  sliding 


i          15                   10 

)  i    >    1    i     <     .     <    i 

Scale 

5                   0        I 

,,,!,,,,        , 

,    ,    ,   .   1    ,   ,    ,   I 
10               5                 0 
Vernier 

FIG.  141. 

scale,  which  is  the  vernier.  The  vernier  is  so  divided  that 
10  divisions  of  its  scale  just  equal  9  divisions  of  the  graduated 
scale.  If  the  0  mark  on  the  vernier  coincides  with  a  division, 
say,  the  0  division  as  in  Fig.  141,  of  the  graduated  scale, 
then  the  division  1  on  the  vernier  stands  at  0.9  on  the  scale; 
2  on  the  vernier  at  1.8  on  the  scale;  and  so  on  for  the  other 
divisions. 

If  the  vernier  be  moved  along  so  that  one  of  its  divisions, 
as  4  in  Fig.  142,  coincides  with  a  division  of  the  scale,  then  the 


\ 

Scale 

/ 

/    20               15 

\i  1  i  i  i  r  !  i  i 

10               5 

i  !  i  [  i  i  !  i 

,  ,  ,°\  N 

i  i  M 

Mil 

10      ._    5  .           0 
Vernier 

FIG.  142. 

division  on  the  vernier  just  to  the  right  or  left  of  the  coinciding 
division  lacks  0.1  of  a  scale  division  of  coinciding  with  a  scale 
division.  The  next  division  of  the  vernier  to  the  right  or 
left  lacks  0.2  of  a  scale  division  of  coinciding  with  a  scale 
division,  and  so  on.  In  this  case,  the  0  point  on  the  vernier 
is  removed  0.4  of  a  division  to  the  left  of  a  scale  division. 

The  reading  then  in  Fig.  142,  that  is,  the  distance  from  the 
0  division  on  the  scale  to  the  0  division  on  the  vernier,  is  7.4. 
If  the  scale  division  is  tenths  of  an  inch,  then  the  reading  is 
0.74  in. 


192 


PRACTICAL  MA  THEM  A  TICK 


If  the  vernier  is  moved  to  the  left  so  that  6  on  the  vernier 
coincides  with  a  division  on  the  scale,  then  0  on  the  vernier 
is  O.G  of  a  scale  division  to  the  left  of  a  scale  division. 

It  is  evident  that  any  number  of  divisions  on  a  scale  could 
be  equal  to  one  greater  number  of  divisions  on  a  vernier,  and 
the  readings  could  be  made  in  a  similar  way.  For  instance, 
in  instruments  for  measuring  angles,  if  the  scale  divisions  are 
to  i°,  then  a  vernier  with  30  divisions  equaling  29  divisions  of 
the  scale  will  give  a  reading  to  ^V  of  \°  or  1'  of  angle. 

157.  Micrometer  with  vernier. — A  micrometer  that  reads 
to  thousandths  of  an  inch  may  be  made  to  read  to  ten-thou- 
sandths of  an  inch  by  putting  a  vernier  on  the  barrel,  so  that 
10  divisions  on  the  vernier  correspond  to  9  divisions  on  the 


Fio.   143. 

thimble.  There  are  eleven  parallel  lines  on  the  sleeve  occu- 
pying the  same  space  as  ten  lines  on  the  thimble.  These  lines 
are  numbered  0,  1,  2,  3,  4,  5,  6,  7,  8,  9,  0.  The  difference 
between  one  of  the  ten  spaces  on  the  sleeve  and  one  of  the  nine 
spaces  on  the  thimble  is  ^V  of  a  space  on  the  thimble  or  m-Jinr 
inch  in  the  micrometer  reading. 

In  Fig.  143(6),  the  third  line  from  0  on  the  thimble  coincides 
with  the  first  line  on  the  sleeve.  The  next  two  lines  do  not 
coincide  by  -fa  of  a  space  on  the  thimble,  the  next  two  marked 
5  and  2  are  i\  of  a  space  apart,  and  so  on.  When  the  mi- 
crometer is  opened  the  thimble  is  turned  to  the  left  and  each 
space  on  the  thimble  represents  mVu  inch.  Therefore,  when 
the  thimble  is  turned  so  that  the  lines  5  and  2  coincide  the 
micrometer  is  opened  Vo  of  TuVa  inch  or  njXniTf  inch.  If  the 
thimble  be  turned  further,  so  that  the  line  10  coincides  with 


GRAPHICAL  METHODS  193 

the  line  7  on  the  sleeve  as  in  (c) ,  the  micrometer  has  been  opened 

TTnhnj-  inch- 
To  read  a  micrometer  graduated  to  ten-thousandths,  note 
the  thousandths  as  usual,  then  observe  the  number  of  divisions 
on  the  vernier  until  a  line  is  reached  which  coincides  with  a 
line  on  the  thimble.     If  it  is  the  second  line  marked  1,  add 
nrfanr;  if  the  third  marked  2,  add  T-jHhnr>  etc. 
As  an  exercise  give  the  different  readings  shown  in  the  figure. 


CHAPTER  XIV 
PRISMS 

158.  Definitions. — Two  planes  are  said  to  be  parallel  when 
they  will  not  meet  however  far  they  be  extended.  That  is, 
they  are  everywhere  the  same  distance  apart. 

A  line  is  parallel  to  a  plane  when  it  will  not  meet  the  piano 
however  far  it  may  be  extended. 

A  line  is  perpendicular  to  a  plane  when  it  is  perpendicular 
to  every  line  of  the  plane  that  passes  through  its  foot. 

A  prism  is  a  solid  whose  ends,  or  bases,  are  parallel  polygons, 
and  whose  sides,  or  faces,  are  parallelograms. 

Fig.  144  is  a  prism.  The  bases  ABCI) 
and  EFGH  are  parallel  polygons.  The 
faces  AEFB,  BFGC,  etc.,  are  parallelo- 
grams. The  lines  AB,  BC,  etc.,  and 
EF ,  FG,  etc.,  are  base  edges.  The  lines 
AE,  BF,  etc.,  are  lateral  edges. 

If   the   lateral    edges    are    per- 
pendicular   to    the    bases   of   the 
prism,  the  prism  is  a  right  prism. 
If  the  lateral  edges  are  not  per- 
pendicular to  the  bases,  the  prism 
is  an  oblique  prism. 
In  an  oblique  prism,  the  faces  are  parallelograms;  but,  in  a 
right  prism,  they  are  rectangles. 

A  prism  is  called  triangular,  square,  rectangular,  hexagonal, 
etc.;  according  as  its  bases  are  triangles,  squares,  rectangles, 
hexagons,  etc.  Fig.  145  is  a  right  triangular  prism. 

A  cross  section  of  a  prism  is  a  section  that  is  perpendicular 
to  the  edges  of  the  prism. 

In  Fig.  145,  MNO  or  either  base  is  a  cross  section. 

The  sum  of  the  edges  of  the  base  is  the  perimeter  of  the  base. 

194 


FIG.  144. 


PRISMS 


195 


The  altitude  of  a  prism  is  the  perpendicular  line  between 
the  two  bases,  as  mn,  Fig.  144.  In  a  right  prism,  the  altitude  is 
the  same  length  as  an  edge.  In  an  oblique  prism,  this  is  not 
true. 

A  right  prism  that  has  rectangles  for  bases  is  called  a  rec- 
tangular solid.  The  cube  is  a  rectangular  solid  all  of  whose 
six  faces  are  squares. 

159.  Surfaces. — The  right  prisms  are  the  forms  of  prisms 
that  are  met  with  usually  in  practical  work.  They  are  the 
ones  considered  here. 

The  lateral  area  of  a  right  prism  is  the  area  of  its  faces,  not 
including    the    two    bases.     Since    the    faces    are    rectangles 
their  areas  can  be  easily  found.     As 
the  area  of  each  face  is  the  product 
of  its  base  by  its  altitude,  the  sum 
of    the    areas    of   the    faces,    or   the 
lateral  area  of  the  prism  is  given  by 
the  following: 

RULE.  The  lateral  area  of  a  right 
prism  equals  the  perimeter  of  the  base 
times  the  altitude. 

The  total  area  of  the  prism  equals 
the  lateral  area  plus  the  area  of  the 
two  bases. 

Since  the  cube  has  six  equal  square  faces,  the  total  area  is 
six  times  the  square  of  an  edge. 

If  S  stands  for  lateral  area,  T  for  total  area,  A  for  area 
of  each  base,  p  for  perimeter  of  the  base,  h  for  altitude, 
and  a  for  an  edge,  the  rules  are  stated  in  the  following 
formulas : 

[33]  S=ph. 

[34]  T  =  ph+2A. 

[35]  T  =  6a2,  for  the  cube. 

[36]  p  =  S-f-h. 

[37]  h  =  S^p. 

Example.  Find  the  total  area  of  a  right  triangular  prism, 
of  altitude  20  ft.,  if  the  edges  of  the  base  are  2  ft.,  3  ft.,  and 
4ft. 


o 


JC 


FIG.   145. 


196  PRACTICAL  MATHEMATICS 

Solution.     p  =  2+3+4  =  9. 

By  [7],  A  =  Vs(s-a)  (s-6)  (s-c). 


s— c=4J  —  4  =  £. 
.'.  A  =  -V/475X 2.5 XI. 5X0.5  =  \/K.4375  =  2.9047. 

By  [34],  r  =  pft+2A. 

/.  7T  =  9X20+2X2.9047=  185.81 -ft.2  Ans. 
160.  Volumes. — In  Fig.  146,  there  are  as  many  cubic  inches 
on  the  base  ABCD  as  there  are  square  inches  in  its  area,  and 
since  there  are  as  many  layers  of  cubic  inches  in  the  rectangular 

H  a 


A 

/ 

/ 

/ 

z 

/ 

/    / 

A  \ 

/ 

/ 

/ 

/ 

/ 

/ 

/    / 

/'  -i 

/ 

/ 

/ 

/ 

2 

/    /  / 

'/<  ' 

k  i    i 
'1    \'\ 

v/ 

•/• 

'•  !,' 

f 

/ 

/ 
/  / 

•/I 

£i-; 

•'!  !' 

1  .' 

DA 

7T& 

7 

r 

—f 

/ 
1  >  / 

/i-- 

-/•-- 

"/"     " 

-/  

-t  

-f  

y-  — 

B 


FIG.  146. 


solid  as  there  are  inches  in  the  altitude  AE,  the  total  number 
of  cubic  inches  in  the  solid  is  found  by  multiplying  the  number 
of  square  inches  in  the  base  by  the  number  of  linear  inches  in 
the  altitude.  Any  right  prism  can  be  taken  in  the  same  man- 
ner as  the  rectangular  solid;  hence  the  following: 

RULE.     The  volume  of  a  right  prism  equals  the  area  of  the 
base  times  the  altitude. 

If  the  prism  is  a  rectangular  solid  this  rule  becomes: 
RULE.     The  volume  of  a  rectangular  solid  equals  the  con- 
tinued product  of  the  length,  breadth,  and  height. 

If  V  stands  for  volume,  and  the  other  letters  as  before,  we 
have  these  formulas: 

[38]     V  =  Ah,  for  any  prism. 
[39]    V  =  a8,  for  the  cube. 
[40]    h  =  V-^A. 
[41]     A  =  Vn-h. 


PRISMS 


197 


Example  1.     One  of  the  concrete  pillars  to  hold  up  a  floor 
in  a  concrete  building  has  a  cross  section  that  is  a  regular 

hexagon.     The  dimensions  are  as  shown  in  Fig.        

147.     Find  its  weight  if  concrete  weighs  138  Ib. 
per  cubic  foot. 

Solution.  The  volume  is  found  by  formula 
[38]  V  =  Ah,  where  h  is  12  ft.  and  A  the  area 
of  a  hexagon  with  one  side  8  in. 


A=6X42Xl.732in.2  = 


6X42X1.732 


V  = 


144 

12X6X42X1.732 


ft.' 


144 


ft.1 


1 

\ 

.. 

w  .  ,  ,      12X6X42X1.732X138     1010  „ 

Weight  =  —  -  =  19121b. 

144 

Example  2.  How  deep  is  a  cistern  in  the  form 
of  a  hexagonal  prism  to  hold  100  bbl.  if  the  base 
is  4  ft.  on  an  edge? 

Solution.     By  formula  [40],  h—V-r-A. 

F=100  bbl.  =  100X31^X231  cu.  in. 
A  =  area  of  hexagon  with  edge  of  4  ft. 
Alt.  of  one  triangle  of  hexagon  =  2X1.  732 

=  3.464  ft, 

Area  of  hexagon  =  6X|(4X  3.464)  =41  .568 
sq.  ft. 
41.568  sq.  ft.  =41.568X144  sq.  in. 


FIG.  147. 


EXERCISES  46 

1.  Find  the  volume  of  the  following  rectangular  solids: 

(a)  8  ft.  by  11  ft.  by  32  ft.  Ans.  2816  ft.8 

(b)  3  ft.  by  4|  ft.  by  7f  ft.  Ans.   104f  ft.3 

(c)  30  ft.  6  in.  by  41  ft.  6  in.  by  12  ft.  Ans.   15,189ft.3 

(d)  17  ft.  2  in.  by  19  ft.  3  in.  by  3  ft.  7  in.  to  nearest  0.001  ft.3 

Ans.   1184.142+  ft.3 

(e)  3  ft.  4  in.  by  14  ft.  7  in.  by  11  ft.  11  in.  to  nearest  0.001  ft.3 

Ans.  579.282+  ft.3 

2.  If  the  inside  dimension  of  a  cubical  tank  is  4  ft.,  find  the  number 
of  barrels  it  will  hold  when  full.  Ans.   15.2  —  bbl. 

3.  Find  the  number  of  cubic  yards  of  earth  to  be  excavated  in  digging 
a  cellar  40  ft.  by  26  ft.  by  7  ft.  Ans.  269.63  -. 


198 


PRACTICAL  MATHEMATICS 


FIG.  148. 


4.  A  rectangular  solid  has  the  three  dimensions:  5.25  ft.,  17.23  ft,, 
and  4.062  ft.     Find  its  capacity  in  bushels.  Ant.  295.3  —  bu. 

5.  Find  the  weight  of  a  block  of  granite  (specific  gravity  2.9)  6  ft. 
by  7  ft.  by  1 J  ft.  An*.  11,418}  Ib. 

6.  Find  the  weight  of  a  solid  cast-iron  pillar  in  the  form  of  a  hexa- 
gonal prism  12  ft.  high  and  3  in.  on  edge  of  base. 

An*.  875  Ib.  nearly. 

7.  A  common  brick  is  2  in.  by  4  in.  by  8  in.     Find  the  number  of 
bricks  in  a  pile  8J  ft.  by  4  ft.  by  10  ft.     (Use  cancellation.)     Ans.  9180. 

8.  How  many  rectangular  solids  3  in.  X4  in.  X9  in.  will  fill  a  box  4  ft.  X 
6ft.X8ft?  Ana.  3072. 

9.  One  hundred  eighty  square 
feet    of   zinc   are   required   for 
lining    the    bottom    and    sides 
of  a  cubical  vessel.     How  many 
cubic  feet  of  water  will  it  hold?  Ans.  216. 

10.  A  box  car  that  is  36 f  ft.  long  and  8$  ft.  wide,  inside  measurements, 
can  be  filled  with  wheat  to  a  height  of  5J  ft.     Find  how  many  bushels 
of  wheat  it  will  hold  if  f  cu.  ft.  are  1  bushel.  Ans.  1317s. 

11.  How  many  cubic  yards  of  soil  will  it  take  to  fill  in  a  lot  50  ft.  by 
100  ft.,  if  it  is  to  be  raised  3  ft.  in  the  rear  end  and  gradually  sloped  to 
the  front  where  it  is  to  be  1J  ft.  deep?  Ans.  416 J. 

Suggestion.  The  vertical  cross  section  the  long 
way  of  the  lot  is  a  trapezoid  having  parallel  sides  *"t~|  fx. 

of    lengths    3   ft.    and    \\  ft.    respectively,   and    an 
altitude  of  100  ft. 

12.  Find   the  number  of  cubic   yards   of   crushed     | 
rock  to  make  a  road  one  mile  in  length  and  of  cross    5] 
section  as  shown  in  Fig.  148. 

Solution.  The  area  of  the  vertical  cross  section 
can  be  found  by  considering  it  as  two  trapezoids  each 
having  parallel  sides  of  8  in.  and  1  ft.  respectively, 
and  altitudes  of  10  ft.  u  -2  * 

Area  of  cross  section  =  (§-f  1)10  =  16 J  sq.  ft. 

No.  of  cu.  ft.  of  rock  =5280X16?  =88,000. 

No.  of  cu.  yd.  of  rock  =88,000-4-27  =  3259^:.     Any. 

13.  One  cubic  inch  of  steel  weighs  0.29  Ib.     An  I-beam  has  a  cross 
section  as  shown  in  Fig.  149,  and  a  length  of  22  ft.     Find  its  weight. 

Ans.  1531+  Ib. 

14.  Find  the  weight  of  steel  beams  10  ft.  in  length  and  of  the  cross 
sections  given  in  Fig.  40,  page  121. 

Am.  821  Ib.;  209.5  Ib.;  300.2  Ib. 

15.  Find  what  weight  of  load  will  be  required  to  cover  a  surface  48  ft. 
long  and  32  ft.  wide,  with  lead  fa  in.  thick,  allowing  5%  of  weight  for 
joints.  Ann.  4  tons  nearly. 


Fiu.  149. 


PRISMS  199 

16.  A  rectangular  box  is  made  with  |-in.  sheet  steel.     Find  weight 
of  box  (no  allowance  for  corners)  if  it  is  4  ft.  by  3.4  ft.  by  2.6  ft. 
(leu.  in.  =  0.29  lb.).  Ans.  342.8  Ib. 

17.  What  length  must  be  cut  from  a  bar  of  steel  ^  in.  by  1|  in.  in 
cross  section  in  order  to  make  1  cu.  ft.?  Ans.  230.4  ft. 

18.  The  base  of  a  right  prism  is  a  triangle  whose  sides  are  12  ft., 
15  ft.,  and  17  ft.,  and  its  altitude  is  8|  ft.     Find  the  lateral  area. 

Ans.  374  ft.2 

19.  Find  the  volume  of  the  above  prism.  Ans.  745.872—  ft.3 
Suggestion.     Use  formula  [7]  to  find  the  area  of  the  ba-se. 

20.  Find  the  volume  of  a  cube  whose  diagonal  is  8  in. 

Ans.  98.534+  in.3 

21.  The  cost  of  digging  a  ditch,  including  all  expenses  and  profits,  is 
estimated  at  27  cents  per  cubic  yard.     Find  the  cost  of  digging  a  ditch 
15  miles  long,    10  ft.  wide  at  the  bottom,  20  ft.  at  the  top,  and  6  ft. 
deep.  Ans.  $71,280. 

22.  A  river  is  76  ft.  wide  and  12  ft.  deep  and  flows  at  the  rate  of  3| 
miles  per  hour.     How  many  cubic  feet  of  water  per  minute  passes  a 
given  point  on  the  river?  Ans.  280,896. 

Suggestion.  The  flow  per  minute  is  the  volume  of  a  prism  of  water  of 
the  cross  section  of  the  river  and  in  length  equal  to  the  distance  the  water 
flows  per  minute. 

23.  A  cistern  7  ft.  long  and  4  ft.  broad  holds  35  bbl.  when  three-fifths 
full;  find  the  depth  of  the  cistern.     Find  the  cost  of  lining  the  bottom 
and  sides  with  zinc  at  5  cents  per  square  foot. 

Ans.  8  ft.  9.27+  in.;  $11.05. 

24.  An  iron  casting  shrinks  about  |  in.  per  linear  foot  in  cooling  down 
to  70°  Fahrenheit.     What  is  the  shrinkage  per  cubic  foot: 

Ans.  53.44—  in.3 

Suggestion.  The  shrinkage  per  cubic  foot  is  the  difference  between  a 
cubic  foot  and  the  volume  of  a  cube  11|  in.  on  an  edge. 

25.  A  fall  of  |  in.  of  rain  is  how  many  barrels  per  square  rod? 

Ans.  4f. 

26.  A  flow  of  300  gallons  per  second  will  supply  water  for  how  deep  a 
stream,  if  the  stream  is  4  ft.  broad  and  flows  5  miles  per  hour? 

Ans.  16yf  in. 
Suggestion.     The  computation  in  the  form  of  cancellation  is 

300X231X60X60 

.  —  -I  t>  3  2  • 


4X5X5280X12X12 

27.  Find  cost  at  40  cents  a  pound  for  sheet  copper  to  line  bottom 
and  sides  of  a  cubical  vessel  7  ft.  on  an  edge,  if  the  sheet  copper  weighs 
12  oz.  per  square  foot.     How  many  barrels  will  the  vessel  hold? 

Ana.  $73.50;  81.45+. 

28.  A  stream,  flowing  5  miles  per  hour,  must  be  of  how  large  cross 
section  to  supply  1?  in.  in  depth  of  water  per  week  for  160  acres  of 
land?  Ans.  28f  in.2 


200  PRACTICAL  MATHEMATICS 

29.  If  a  glass  rod  1  in.  long  at  0°  centigrade  is  increased  to  1.000008  in. 
at  1°  centigrade,  find  the  increase  in  volume  of  a  cubic  inch  of  glass 
when  heated  from  0°  to  1°  centigrade.  Ant.  0.000024+  in.1 

STONEWORK 

161.  Terms. — A    full    description    cannot   be   given    here. 
For  such  the  student  is  referred  to  a  trade  handbook. 

Stonework  where  the  stones  are  broken  with  the  hammer 
only  is  called  rubble-work.  If  the  stones  are  laid  in  courses  it 
is  called  coursed  rubble.  When  the  stones  showing  in  the  out- 
side face  of  the  wall  are  squared,  the  work  is  designated  as 
ashlar.  If  all  the  stones  of  a  course  are  of  the  same  height,  the 
work  is  called  coursed  ashlar.  When  the  stones  are  of  dif- 
ferent heights  it  is  called  broken  ashlar.  Ashlar  work  is  both 
hammer-dressed  and  chisel-dressed.  Any  stonework  where 
any  other  tool  than  the  hammer  is  used  for  dressing  is  called 
cut- work. 

162.  Estimating    cost    of    stonework. — In    estimating   the 
cost  of  stonework,  the  custom  varies  greatly;  we  find  it  vary- 
ing even  among  the  contractors  of  the  same  city.     Cut-work 
is  often  measured  by  the  number  of  square  feet  in  the  face  of 
the  wall. 

Rubble- work  is  almost  universally  measured  by  the  perch, 
but  the  pereh  used  varies  greatly.  The  legal  perch  of  24f 
cu.  ft.  is  seldom  used  by  stone  masons.  The  perch  of  16$ 
cu.  ft.  is  the  one  most  used.  That  of  25  cu.  ft.  or  of  22  cu.  ft. 
is  sometimes  used. 

Stonework  on  railroads  is  usually  measured  by  the  cubic 
yard. 

Openings,  as  a  rule,  are  not  deducted  if  containing  less  than 
70  sq.  ft. 

EXERCISES  47 

1.  Find  the  cost  of  laying  a  hammer-dressed  ashlar  wall,  45  ft.  long, 
6  ft.  high  and  2  ft.  thick,  at  $2.75  a  perch,  using  the  22-cu.  ft.  perch. 

Ana.  $67.50. 

2.  Find  the  cost  of  making  a  rubble-work  wall  at  $3.25  a  perch,  includ- 
ing all  the  material,  under  a  building  25  ft.  by  60  ft.,  the  wall  to  be  30  in. 
thick  and  8  ft.  high.     Use  the  16J-cu.ft.  perch.  Ans.  $630.30. 

3.  Find  the  cost  of  laying  the  stonework  in  two  abutments  for  a  bridge, 
each  abutment  to  be  8  ft.  high,  3J  ft.  thick,  20  ft.  long  at  the  bottom  and 


PRISMS  201 

15  ft.  at  the  top.     (The  shape  is  a  trapezoid.)     The  price  for  laying  is 
$1.75  per  cubic  yard.  Ans.  $63.52. 

4.  Find  the  cost  of  building  a  sandstone  rubble-work  wall  for  a  base- 
ment 36  ft.  by  47  ft.,  8  ft.  high,  and  the  wall  18  in.  thick.  (Use  outside 
measurements  and  make  no  allowances  for  openings.)  The  stone  costs 
$1.40  a  perch  (25  cu.  ft.),  and  the  labor  of  laying,  including  lime  and 
sand,  is  $1.25  a  perch.  Ans.  $211.15. 

BRICKWORK 

163.  Brick. — The    size    of    brick    varies.     In    the    United 
States,    there  is   no  legal  standard.     The  common  brick  is 
approximately  given  as  8  in.  by  4  in.  by  2  in.     In  the  New 
England  States,  they  average  about  7f  in.  by  3f  in.  by  2j  in. 
In  most  of  the  western  states,  the  common  brick  averages  about 
8|  in.  by  4 1  in.  by  2^  in.     The  brick  from  the  same  lot  may 
vary  as  much  as  -fo  in.,  depending  upon  the  degree  to  which 
they  are  burnt.     The  hard-burned  bricks  are  the  smaller. 

Walls  made  from  these  bricks  are  about  9,  13,  18,  and  22 
inches  in  thickness,  that  is,  1,  1^,  2,  and  1\  bricks. 

Pressed  bricks  are  usually  larger  than  common  bricks;  the 
prevailing  size  is  8f  in.  by  4|  in.  by  2f  in. 

164.  Estimating    number,    and   cost   of   brickwork. — This 
cannot  be  discussed  in  full  here.     The  student  is  referred  to 
an  architect's  and  builder's  handbook  for  the  details.     Plain 
walls  are  quite  universally  figured  at  15  bricks  to  the  square 
foot,  outside  measure,  of  8-  or  9-in.  wall;  22^  bricks  per  square 
foot  of  12-  or  13-in.  wall;  30  bricks  per  square  foot  of  16-, 
17-,  or  18-in.  wall;  and  1\  bricks  for  each  additional  4  or  4|  in. 
in  thickness  of  wall.     These  figures  are  used  without  regard  to 
the  size  of  the  bricks,  the  effect  of  the  latter  being  taken  into 
account  in  fixing  the  price  per  thousand.     No  deduction  is  made 
for  openings  of  less  than  80  sq.  ft.,  and  when  deductions  are 
made  for  larger  openings  the  width  is  measured  2  ft.  less  than 
the  actual  width.     Hollow  walls  are  measured  as  if  solid. 
For.  chimney-breasts,  pilasters,  detached  chimneys,  and  other 
forms  the  student  is  referred  to  a  builder's  handbook. 

Example.  Find  the  cost  of  brickwork  in  the  walls  of  a 
house  26  ft.  by  34  ft.,  no  cross  walls,  the  basement  walls  to  be 
13  in.  thick;  the  first  story  walls,  13  in.  thick;  second  story 
walls,  9  in.  thick;  height  of  basement  walls  to  top  of  first  floor 


202  PRACTICAL  MATHEMATICS 

joists,  9  ft.;  from  first  floor  joists  to  top  of  second  floor  joists, 
10  ft.  6  in.;  from  second  floor  joists  to  plate,  9  ft.  (The 
chimneys,  openings,  and  pressed  brickwork  are  not  considered 
as  the  method  of  estimating  has  not  been  given.)  The  cost 
of  brick  and  laying,  including  lime,  sand,  scaffolding,  etc., 
is  $10  per  thousand. 

Solution.     Basement  and  first  story  walls: 
Girth  of  house  =  2X26  ft. +2X34  ft.  =  120  ft. 
Height  of  wall  =  9  ft. +  10  ft.  6  in.  =  19£  ft. 
Thickness  of  wall  is  13  in.  and  hence  22£  brick  are  counted  per 
square  foot.     Hence  the  number  of  bricks  required  is 

120  X  19£  X22£  =  52,650. 
Similarly  for  the  second  story, 

120X9X15=16,200. 

/.total  number  of  bricks  =  52,650+ 16,200  =  68,850. 
.'.cost  =  $10X68.850  =  $688.50.  Ans. 

EXERCISES  48 

1.  Find  the  cost  at  $9.50  per  thousand  to  cover  brick,  material,  and 
labor  to  build  a  brick  wall  on  the  front  and  one  side  of  a  corner  lot  50  ft. 
by  100  ft. ;  the  wall  to  be  6J  ft.  high  and  a  brick  and  a  half  thick,  allow- 
ance being  made  for  one  opening  12  ft.  in  length  (count  it  2  ft.  leas). 

Ans.  $194.51. 

2.  The  following  is  about  the  cost  of  furnishing  and  laying  1500  bricks, 
or  one  day's  work : 

1500  bricks  at  $6.00  per  M, 
~~  1  barrel  of  lime  at  $1.00, 

?  9  bushels  of  sand  at  5  cents, 


x    *—  1  day's  work  for  mason  at  $5.76, 

!  1  day's  work  for  helper  at  $3.32. 


— 4'2tf ^  Using  this  as  a  basis  for  estimating,  find  the  cost 

of  building  the  walls  of  an  apartment  house  25 
Fio.  160.  ffc  by  54  ft  ^  41  ft  nign  jn  front  ftnci  36  ft   ^  tne 

rear.     The  walls  are  to  be   13  in.  thick  and  no  allowances  for  open- 
ings. Ans.  $1759.20. 

3.  Find  the  cost  of  common  brick  in  the  pier  with  a  cross  section  as 
shown  in  Fig.  150,  and  a  height  of  12  ft.  6  in.,  at  $7.00  per  thousand. 
Count  20  brick  to  1  cu.  ft.  Ans.  $24.15. 

4.  How  many  enamel  brick  4  in.  by  8  in.  are  required  to  face  a  wall 
30  ft.  long  and  12  ft.  high,  deducting  for  two  windows  4  ft.  by  8  ft.  and 
one  door  3ft.  4  in.  by  10ft.?  Ans.  1182. 


CHAPTER  XV 
CYLINDERS 

165.  Definitions. — A  right  circular  cylinder,  or  a  cylinder 
of  revolution,  is  a  solid  formed  by  revolving  a  rectangle 
about  one  of  its  sides  as  an  axis. 

Thus,  in  Fig.  151,  the  rectangle  OABO'  is  revolved  about  OO'  as  an 
axis.  This  forms  the  cylinder  as  drawn. 

From  this  definition,  the  two  bases  are  circles,  and  the 
lateral  surface  is  a  curved  surface.  The  axis  of  the  cylinder 
is  the  line  00'  joining  the  centers  of  the  bases.  It  is  per- 

C' 


FIG.   151. 

pendicular  to  the  bases,  and  hence  it  is  equal  to  the  altitude 
of  the  cylinder.  The  cross  section  of  a  cylinder  is  a  section 
perpendicular  to  the  axis. 

A  cylinder  is  inscribed  in  a  prism  when  its  bases  are  in- 
scribed in  the  bases  of  the  prism  (see  Fig.  152).  The  prism 
is  then  circumscribed  about  the  cylinder.  A  cylinder  is  cir- 
cumscribed about  a  prism,  or  the  prism  is  inscribed  in  the 
cylinder,  when  the  bases  of  the  prism  are  inscribed  in  the 
bases  of  the  cylinder. 

166.  Area  and  volume. — If  the  lateral  surface  of  a  right 
cylinder  could  be  peeled  off  and  spread  out,  it  would  form 
a  rectangle  of  width  equal  to  the  altitude  of  the  cylinder, 

203 


204  PRACTICAL  MAT  HEMATICS 

and  of  length  equal  to  the  circumference  of  the  base  of  the 
cylinder.  From  this  we  get  the  following: 

RULE.  The  area  of  the  lateral  surface  of  a  right  cylinder 
equals  the  circumference  of  the  base  times  the  altitude. 

The  total  area  equals  the  lateral  area  plus  the  area  of  the  two 
bases. 

From  a  consideration  similar  to  that  for  the  prism,  Art.  160, 
the  volume  of  the  cylinder  is  obtained  by  the  following: 

RULE.  The  volume  of  a  cylinder  equals  the  area  of  the  base 
times  the  altitude. 

The  similarity  of  the  cylinder  and  the  prism  is  seen  if  the 
cylinder  is  thought  of  as  a  prism  having  a  very  great  number 
of  sides  to  the  base. 

The  above  rules  hold  when  the  cylinder  is  not  circular. 

Since  the  altitude  times  the  cross  section  gives  the  volume, 
the  altitude  equals  the  volume  divided  by  the  area  of  the  base 
or  cross  section.  Also  the  area  of  the  base  equals  the  volume 
divided  by  the  altitude. 

If  S  stands  for  lateral  area,  T  for  total  area,  A  for  area  of 
base,  and  h  for  altitude,  the  rules  given  are  stated  in  the 
following  formulas: 

[42]  S  =  Ch 

[43]  V  =  Ah 

[44]  h=V-rA  =  V^-irr2. 

[45]  A  =  V-hh. 

167.  The  hollow  cylinder. — The  volume  of  a  hollow  cylinder 
may  be  found  by  subtracting  the  volume  of  the  cylindrical 
hollow  from  the  volume  of  the  whole  cylinder.  If  R  is  the 
radius  of  the  cylinder,  and  r  the  radius  of  the  hollow,  then  the 
volume  of  the  hollow  cylinder  is  as  follows: 

[46]  V  =  7rR%-7TT2h  =  Th(R2-r2)-7rh(R+r)(R-r). 

Example  1.  Find  the  number  of  cubic  inches  of  copper  in  a 
hollow  cylinder  7  in.  long,  inner  diameter  6  in.  and  outer 
diameter  8  in. 

Solution.     By  [46] ,  V  =  irh(R+r)(R-r). 

/.  F  =  3.1416X7(4+3)(4-3)  =  153.938  in.8 

Example  2.  Find  the  effective  heating  surface  of  a  boiler 
of  diameter  5  ft.  and  length  16  ft.;  with  54  tubes  3|  in.  in 


CYLINDERS 


205 


diameter,  assuming  that  the  effective  heating  surface  of  the 
shell  is  one-half  the  total  surface. 

Solution.     By  [42],  effective  heating  surface  of  the  shell  = 


145.299  ft.2 


Effective  heating  surface  of  54  tubes 


/.total  heating  surf  ace  =  145.299  +  791.  683  =  937  ft.2  nearly. 

Example  3.  What  is  the  weight  of  a  cylindrical  shaft 
of  marble  3  ft.  in  circumference  and  9  ft.  high? 

Solution.     r  =  f  (3  -f-3.1416)  =0.47746. 

By  [43],  F  =  7rr2/i  =  3.1416X0.477462X9  =  6.4457  ft,3 

Weight  =  2.7  X  62.5  X  6.4457  =  1087  Ib. 

Since  62.5  Ib.  =  weight  of  1  cu.  ft.  of  water,  and  2.7  =  specific 
gravity  of  marble. 


EXERCISES  49 

1.  If  the  radius  of  the  base  of  a  right  circular  cylinder  is  5  in.  and 
the  altitude  is  8  in.,  find  the  lateral  area,  the  total  area,  and  the  volume. 

Ans.  251.33  in.2;  408.4  in.2;  628.3  in.3 

2.  If  r,  d,  h,  A,  S,  and  V  have  the  meanings  given  in  the  preceding 
articles: 

(a)  Given  A  =48  sq.  in.  and  h  =  16  in. ;  find  r,  S,  and  F. 

(b)  Given  F  =  4800  cu.  ft.  and  A  =  160  sq.  ft.;  find  d  and  S. 

Ans.   (a)  r  =  3.909 -in.;  S  =  393.0 -sq.in.;  7  =  768  cu.  in. 

3.  Find  the  area  of  the  rubbing  surface  in  a  steam  cylinder  91J  in.  in 
diameter,  the  stroke  of  the  piston  being  6  ft.  8  in.          Ans.  159.7  —  ft.2 

4.  Find  the  total  area  and  the  volume  of  a  cylinder  whose  radius  is 
7  ft,  and  whose  altitude  is  12  ft.  Ans.  835.66  ft.2;  1847+  ft.3 

5.  Find  the  number  of  barrels  each  of  the  follow- 
ing cylindrical  tanks  will  hold: 

(a)  Diameter  5ft,,  depth    5ft.  Ans.  23.3. 

(b)  Diameter    8  ft,,  depth    9  ft,          Ans.  107.4. 

(c)  Diameter  20  ft.,  depth  19  ft.       Ans.  1417.5. 

(d)  Diameter  30  ft.,  depth  20  ft.        Ans.  3357.3. 
Suggestion.     Use  1  bbl.  =4.211  cu.  ft. 

6.  Find  the  number  of  cubic  yards  of  earth  removed 
in  digging  a  tunnel  175  yd.  long,  if  the  cross  section 
is  a  semicircle  with  a  radius  of  14  ft.        Ans.  5986.5. 

7.  A  peck  measure  is  to  be  made  8  in.  in  diameter. 
it  be?     (See  Fig.  153.) 


FIG.   153. 
How  deep  should 
Ans.  10.695  in. 


206  PRACTICAL  MATHEMATICS 

Suggestion.  1  pk.  -  J  of  2150.42  cu.  in.  =537.605  cu.  HI. 
Area  of  base  -8*  X0.7854  -50.2656  sq.  in. 
Altitude  —  volume  -r-area  of  base. 

8.  A  cylindrical  oil  tank  3  ft.  in  diameter  and  10  ft.  long  will  contain 
how  ninny  gallons  of  oil?  Ans.  528.8—. 

9.  The  external  diameter  of  a  hollow  cast-iron  shaft  is  18  in.,  and 
its  internal  diameter  is    10  in.     Calculate  its  weight  if  the  length  is 
20  ft.,  and  cast  iron  weighs  0.26  Ih.  per  cubic  inch. 

Ans.  10,980  Ib.  nearly. 

10.  Water  is  flowing  at  the  rate  of  10  miles  per  hour  through  a  pipe 
16  in.  in  diameter  into  a  rectangular  reservoir  197  yd.  long  and  87  yd. 
wide.     Calculate  the  time  in  which  the  surface  will  be  raised  3  in. 

Ana.  31.38  minutes. 

10X5280 

Suggestion.  Ten  miles  per  hour  is—  —=880    ft.    per    minute. 

60 

Now  find  the  number  of  cubic  feet  of  water  that  will  flow  through 
the  16-in.  pipe  in  1  minute.  Then  find  the  number  of  cubic  feet 
required  to  fill  the  reservoir  3  in.  The  quotient  found  by  dividing 
the  required  number  of  cubic  feet  by  the  flow  per  minute  is  the  time 
in  minutes. 

11.  In  a  table  giving  weights  and  sizes  of  square  nuts  for  bolts,  a  nut 
2  in.  square  and  li  in.  thick,  with  a  hole  Ifa  in.  in  diameter,  has  a  given 
weight  of  1.042  Ib.     Use  wrought  iron  and  find  this  weight. 

12.  Find  the  length  of  steel  wire  in  a  coil,  if  its  diameter  is  0.025  in. 
and  its  weight  is  50  Ib.     (Use  1  cu.  in.  weighs  0.29  Ib.) 

Ans.  29,270  ft.  nearly. 

13.  A  cylindrical  cistern  is  6  ft.  in  diameter.     Find  the  depth  of  the 
water  when  containing  10  barrels.  .Ixx.  17. S7  in. 

14.  A  nugget  of  gold  is  dropped  into  a  cylinder 
of  water  and  raises  the  surface  of  the  water  li  in. 
If  the  cylinder  is  2  in.  in  diameter  and  25  grains 
of  gold  are  worth  $1,  find  the  value  of  the  nugget. 

Ans.  $767. 

16.  In  a  table  giving  size,  etc.,  of  wrought-iron 
washers,  a  washer  3J  in.  in  diameter  with  a  hole 
1 J  in.  in  diameter  is  A  in.  thick.  Find  the  number 
of  these  in  a  keg  of  200  Ib.  Ans.  582. 

16.  Find  the  per  cent  of  error  in  the  following  FIG.   l.vi. 
rule  which  applies  to  round  bars,  (a)  for  wrought 

iron,  (b)  for  cast  iron.  Multiply  the  square  of  the  diameter  in  inches  by 
the  length  in  feet,  and  that  product  by  2.6.  The  product  will  be  the 
weight  in  pounds  nearly. 

Ans.  (a)  1.5-%  too  small;  (b)  6.1  +  %  too  large. 

17.  A  conduit  made  of  concrete  has  a  cross  section  as  shown  in  Fig. 
154.     How  many  cubic  yards  of  concrete  are  used  in  making  500  yd.  of 
this  conduit?  Ans.  1113.4  yds. 


CYLINDERS 


207 


18.  In  a  table  giving  weights  and  areas  in  cross  section  of  steel  bars,  a 
round  steel  bar  f  in.  in  diameter  has  its  area  given  as  0.1104  in.2  and 
weight  0.376  Ib.  per  linear  foot.     Verify  these  results  if  steel  weighs 
489.6  Ib.  per  cubic  foot. 

19.  A  water  tank  in  a  Pullman  car  has  a  vertical  cross  section  as  shown 
in  Fig.  155,  and  a  length  of  52  in.     Find  its  capacity  in  gallons.     The 
arc  is  a  part  of  a  circle.  Ans.  68.3  gal. 

20.  A  rod  of  copper  8  in.  long  and  1  in.  in  diameter  is  drawn  into  wire 
of  uniform  thickness  and  200  ft.  long.     Find  the  diameter  of  the  wire. 

Ans.  0.0577  in. 

21.  The  rain  which  falls  on  a  house  22  ft.  by  36  ft.  is  conducted  to  a 
cylindrical  cistern  8  ft.  in  diameter.     How  great  a  fall  of  rain  would  it 
take  to  fill  the  cistern  to  a  depth  of  1\  ft.?  Ans.  5.712  in. 

22.  Find  the  weight  of  7  miles  of  |-in.  copper  wire,  if  copper  weighs 
0.319  Ib.  per  cubic  inch.  Ans.   1736.26  Ib. 

23.  Find  the  greatest  tensile  force  a  copper  wire  0.25  in.  in  diameter 
can  stand  without  breaking.     (See  Table  VII.) 

Aris.  Less  than  1472.6  Ib. 


FIG.  155. 


By  tensile  force  is  meant  a  pulling  force.     The  problem  asks  how  great 
a  weight  the  wire  will  hold  when  hanging  vertically. 

24.  Calculate  the  size  of  a  square  wrought-iron  bar  to  stand  a  pull  of 
43,000  Ib.     (See  Table  VII.)  Ans.  0.927  in.  square. 

25.  What  should  be  the  diameter  of  a  round  cast-iron  bar  which  is 
subjected  to  a  tension  of  30,000  Ib.,  if  the  pull  on  each  square  inch  of 
cross  section  is  2400  Ib.?  Ans.  3.989  in. 

26.  Find  the  pull  per  square  inch  necessary  to  break  a  rod  2|  in.  in 
diameter,  which  breaks  with  a  load  of  270,000  Ib. 

Ans.  55,000  Ib.  nearly. 

27.  If  a  wrought-iron  bar  2  in.   by   lj  in.  in  cross  section  breaks 
under  a  load  of  125,000  Ib.,  what  load  will  break  a  wrought-iron  rod 
2£  in.  in  diameter?  Ans.  245,400  Ib.  nearly. 

28.  A  cast-iron  bar  has  an  elliptical  cross  section  with  axes  6  in.  and 
4  in.     Find  the  pull  per  square  inch  of  cross  section  under  a  total  tensile 
load  of  125,000  Ib.  Ans.  6631  Ib. 

29.  A  wrought-iron  cylindrical  rod  2000  ft.  long  and  1^  in.  in  diameter 
is  suspended  vertically  from  its  upper  end.     What  is  the  total  pull  at 
this  end,  and  the  pull  per  square  inch  of  cross  section? 

Ans.   11.87511).;  6720  Ib. 


•JMS 


PRACTICAL  MA  THEM  A  TICS 


1 


Fio.  150. 


30.  Find  the  length  of  a  wrought-iron  bar,  supported  vertically  at  its 
upper  end,  that  will  just  break  under  its  own  weight.    Ans.  15,000  ft. 

31.  About  what  is  the  strength  of  an  18  gage  B.  and  S.  wrought-iron 
wire?     (See  Table  VII.)  An*.  63.8  Ib. 

32.  A  trolley  wire  is  copper  and  00  gage  B.  and  8.     Find  the  pull 
it  will  take  to  break  the  wire.  Ans.  3130  Ib. 

33.  A  cylinder  to  cool  lard  is  4  ft.  in  diameter  and  9  ft.  long  and  makes 
4  R.  P.  M.     As  the  cylinder  revolves,  the  hot  lard  covers  the  surface 
I  in.  deep.     How  many  pounds  of  lard  will  it  cool  in  1  hour  if  the  specific 
gravity  of  lard  is  0.9?  Ans.   15,900  nearly. 

34.  If  a  tank  5  ft.  in  diameter  and  10  ft. 
deep  holds  10,000  Ib.  of  lard,  what  will  be  the 
depth  of  a  tank  of  2000  Ib.  capacity  if  its 
diameter  is  3  ft.  ?  If  this  tank  has  a  jacket 
around  it  on  the  bottom  and  sides  3  in.  from 
the  surface  of  the  tank,  how  many  gallons  of 
water  will  the  space  between  the  jacket  and 
tank  hold?  Ans.  5J  ft.;  124  nearly. 

36.  Find  the  volume  of  a  wash  boiler  if 
the  bottom  is  in  the  form  of  a  rectangle  with 
a  semicircle  at  each  end.  The  rectangle  is  10 
in.  by  14  in.,  and  the  semicircles  are  on  the 
smaller  dimensions.  The  depth  of  the  boiler 
is  16  in.  .4ns.  15.14—  gal. 

36.  Find  the  height  of  a  10-gallon  wash  boiler  whose  base  is  10  in. 
wide  with  semicircular  ends,  the  length  of  the  straight  part  of  the  sides 
being  9i  in.  Ans.  13.5—  in. 

37.  A  certain  handbook  gives  the  following  "rules  of  thumb"  for 
finding  the  volume  in  gallons  of  a  cylindrical  tank: 

(1)  V  (in  gal.)  =  (diameter  in  feet)1  X5J  X  (height  in  feet). 

(2)  V  (in  gal.)  =  (diameter  in  feet)*X§  of  height  in  inches  less  2% 
of  same. 

Find  per  cent  of  error  for  each  rule. 

Ans.  (1)  0.003  +  %  too  small;  (2)  0.08  +  %  too  large. 
Remark.     Rule  (2)  is  a  quick  rule  to  apply  if  the  2%  is  disregarded. 
This  rule  is  in  common  use  by  many  estimators. 

38.  In  making  the  pattern  of  a  teakettle  with  an  elliptical  bottom 
to  hold  6  quarts,  it  is  decided  to  have  the  bottom  an  ellipse  with  axes 
10  in.  and  7  in.     Find  the  height.  Ans.  6.30+  in. 

39.  How  long  a  piece  of  copper  will  it  take  to  make  the  body  of  the 

above  teakettle?     KTse  Cir.   =2»x-J-')  Ans.  26.70  in. 

40.  In  drilling  in  soft  steel,  a  IjViri.  twist-drill  makes  37  R.  P.  M. 
with  a  feed  of  «'„  in.     Find  the  number  of  cubic  inches  cut  away  in  3J 
minute*.  Ans.  3.1+. 

41.  A   j3«-in.    twist-drill  makes  310  R.  P.  M.  with  a  feed  of  T*J  in. 
Find  the  volume  cut  away  in  31  minutes.  Ans.  0.24—  in.3 


CYLINDERS  209 

42.  In  turning  a  steel  shaft  6  in.  in  diameter  and  4  ft.  long,  the  cutting 
speed  was  36  ft.  per  minute  and  the  feed  0.125  in.     Find  the  time  re- 
quired for  turning  the  shaft.     If  the  depth  of  the  cut  was  0.05  in.,  find 
the  amount  of  metal  removed.  Ans.  16 1  min.  nearly;  45.2  in.3 

43.  Find  the  number  of  pounds  of  cast  iron  turned  off  per  hour  in  the 
following:     (Consider  the  cutting  as  if  on  a  plane.) 

Speed  per  min.  Depth  of  cut  Breadth  of  cut 

(a)  37.90  ft.  0.125  in.  0.015  in.  Ans.  13.3. 

(b)  25.82  ft.  0.015  in.  0.125  in.  Ans.  9.06. 

(c)  25.27ft.  0.048  in.  0.048  in.  Ans.  10.9. 

44.  A  steam  chest  cover  is  42  in.  by  24  in.     How  many  steel  studs  1  j  in. 
in  diameter  should  be  used  to  hold  the  cover*  if  the  steam  pressure  is 
160  Ib.  per  square  inch?     The  diameter  of  the  bolts  at  the  bottom  of  the 
thread  is  1.065  in.     Allow  a  stress  of  11,000  Ib.  per  square  inch. 

Ans.   16.5  bolts  or  18  to  make  the  number  even. 

Solution.     42  X 24  X 160  Ib.  =  161,280  Ib.  total  pressure. 

161,280-^11,000  =  14.662  in.2=area  in  cross  section  of  all  bolts. 

0. 7854  Xl.0652  =  0.89082+  in.2=  area  of  one  bolt. 

14.662  -7-0.89082  =  16.5-. 

.'.  number  of  bolts  is  18  to  be  even. 

46.  The  flanges  at  the  joining  of  two  ends  of  flanged  steam  pipes  9  in. 
in  inside  diameter  are  bolted  together  by  12  bolts  \  in.  in  diameter.  If 
the  pressure  in  the  pipes  is  200  Ib  per  square  inch,  find  what  each  bolt 
must  hold.  How  much  is  this  per  square  inch  cross  section  of  the  bolts? 
Suppose  that  bolts  have  10  pitch 
U.  S.  S.  thread.  This  makes  the 
root  diameter  0.620  in.  (See  Fig. 
157.) 

Ans.   1060.3  Ib.;  3512+  Ib. 

46.  As  in  the  last  exercise,  if 

the  steam  pipe  is  18  in.  in  diam-  FIG.  157. 

eter,  and  allowing  the  same  pull 

per  square  inch  of  cross  section  of  each  bolt,  find  the  number  of  bolts 

If  in.   in  diameter  at  a  joint  of  the  pipe.     (A  li-in.  bolt  with  7  pitch 

U.  S.  S.  thread  is  0.940  in.  in  diameter  at  root  of  thread.)  Ans.  22. 

47.  The  following  rule  is  often  used  to  find  the  heating  surface  of  any 
number  of  tubes  in  a  steam  boiler:  Multiply  the  number  of  tubes  by  the 
diameter  of  one  tube  in  inches,  this  product  by  its  length  in  feet,  and  then 
by  0.2618.     The  final  product  is  the  number  of  sq.  ft.  of  heating  surface. 
Using  this  rule,  what  is  the  heating  surface  of  66  3-in.  tubes  each  18  ft. 
long?     Does  the  rule  give  the  correct  result?  Ans.  933  ft.2;  yes. 

48.  To  find  the  water  capacity  of  a  horizontal  tubular  boiler,  find  f 
the  volume  of  the  shell  and  subtract  from  this  the  volume  of  all  the  tubes. 
Find  the  water  capacity  of  a  horizontal  tubular  boiler  185  ft.  long,  66  in. 
ui  diameter,  with  72  3-in.  tubes.  Ans.  227.6+  ft.3 

49.  The  steam  capacity  of  a  horizontal  boiler  is  often  reckoned  as 


210  PRACTICAL  MATHEMATICS 

one-third  the  volume  of  the  shell.     Find  the  steam  capacity  of  a  hori- 
zontal boiler  18  ft.  long  and  78  in.  in  diameter.  Ans.  199  +  ft.J 

50.  Use  the  following  rule  and  find  the  heating  surface  of  a  boiler 
12  ft.  long,  5  ft.  in  diameter,  and  having  52  2J-in.  tubes. 

Ana.  518.7  ft.1  to  550.7  ft,1 

Rule.  In  finding  the  heating  surface  in  a  horizontal  boiler,  it  is  cus- 
tomary to  take  one-half  to  two-thirds  of  the  lateral  area  of  the  shell, 
the  lateral  area  of  the  tubes,  one-half  to  two-thirds  the  area  of  the  ends 
of  the  boiler,  and  subtract  the  areas  of  both  ends  of  the  tubes. 

61.  A  steam  boiler  is  72  in.  in  diameter,  18  ft.  long,  and  contains  70 
tubes  4  in.  in  diameter.     Find  the  heating  surface,  using  one-half  in  the 
rule.  Ans.  1505+  ft.1 

62.  Find  the  steam  capacity  of  a  boiler  4  ft.  in  diameter  and  16  ft. 
long,  if  the  height  of  the  segment  occupied  by  the  steam  is  18 in.?     Is 
this  more  or  less  and  how  much  than  one-third  the  total  capacity  of  the 
boiler  shell?  Ans.  68.9  ft.J;  1.9  ft.1  more. 

Suggestion.     Using    [30(b)l    as    the    most    convenient, 

A  =^X1.52X\/4- -0.608  =  4.305  ft.1 
o  \  l.o 

Volume  =  16  X4.305  ft.3  =  68.9  ft.' 

63.  The  cylinder  of  a  pump  is  6  in.  in  diameter,  the  length  of  stroke 
8  in.,  and  the  number  of  strokes  per  minute  160.     Find  the  flow  in  gallons 
per  minute  if  the  pump  is  double  acting,  that  is,  pumps  the  cylinder  full 
rach  stroke.  Ans.   156.7—  gal. 

64.  When  the  piston  of  a  hand  pump  is  3  in.  in  diameter,  and  the  sup- 
ply of  water  is  drawn  from  a  depth  of  25  ft.,  what  pressure  is  required  on 
the  handle  24  in.  from  the  fulcrum  when  the  piston  rod  is  attached  3i 
in.  from  the  fulcrum?  Ans.   10.30—  Ib. 

66.  The  Cleveland  Twist  Drill  Company  records  a  test  in  which  a 
IJ-in.  "Paragon"  high-speed  drill  removed  70.56  cu.  in.  of  cast  iron  per 

minute.  The  penetration  per 
minute  was  57J  in.,  the  feed  1*0 
in.,  and  the  drill  made  575 
H.  P.  M.  Do  these  numbers 


agree 


p        ,K  66.  What    will  be  the  weight 

of     a     cast-iron    pipe    10    ft. 

long,  2  ft.  in  outer  diameter,  and   1  in.  thick?     (Use  0.26  Ib.  per  cubic 
inch.)  Ans.  2254  Ib. 

67.  A  tank  car  with  a  cylindrical  tank  8  ft.  in  diameter  and  34  ft. 
long  will  hold  how  many  gallons?     What  weight  of  oil  will  it  hold  if  the 
specific  gravity  of  oil  is  0.94?  Ans.   12,784  gal.;  100,400  Ib. 

68.  Find  the  height  of  a  cylindrical  tank  having  a  diameter  of  30  in. 
in  order  that  it  may  hold  4  barrels.  Ans.  41 A  in. 

69.  If  a  bar  1J  in.  in  diameter  weighs  6.01  Ib.  per  foot  of  length, 
what  i»  the  weight  per  foot  of  a  bar  1 J  in.  square  and  of  the  same  material? 

Ans.  7.65+  Ib. 


CYLINDERS  211 

60.  Find  the  weight  of  a  hollow  hexagonal  bar  16  ft.  long  and  weighing 
0.28  Ib.  per  cubic  inch.     The  cross  section  is  a  regular  hexagon  1  \  in.  on 
a  side,  with  a  circle  \\  in.  in  diameter,  at  the  center.     (See  Fig.  158.) 

Ans.  152ilb. 

61.  A  cylindrical  tank  22  ft.  long  and  6  ft.  in  diameter  rests  on  its  side 
in  a  horizontal  position.     Find  the  number  of  gallons  of  oil  it  will  hold 
when  the  depth  of  the  oil  is  8  in.     When  1  ft.  6  in.     When  2  ft.  6  in. 
Use  formula  [30(b)l  for  finding  the  area  of  the  segment. 

Ans.  282.5  gal.;  909.3  gal. 

62.  The  segment  in  Fig.  159  is  a  counter-balance  5|  in.  thick.     Find 
its  weight  if  made  of  cast  iron  weighing  0.26  Ib.  per  cubic  inch. 

Solution.     Area   of   segment   AnB=are&   of   sector   AOBn  —  area   of 
triangle  AOB. 

Area  of  sector  AO£n  =  |X842XO.  7854  =  923.63  in.2 
Area  of  triangle  AOB  =  |X42X21  X\/3  =  763.81  in.2 
Area  of  segment  AnB  =923.63  in.2-763.81  in.2  =  159.82  in.2 
Volume  of  counter-balance  =5^X159.82  in.3  =879.0  in.3 
Weight  of  counter-balance  =879.0X0.26  Ib.  =228.5  Ib.     Ans. 

63.  What  is  the  diameter  of  a  single  round 
rod  in  order  that  it  may  be  as  strong  as  three 
rods   having   diameters   of    ^  in.,    1  in.,   and 
1J  in.  respectively?  Ans.   1.87+  in. 

64.  Because  the  body  of  a  bolt  is  greater  in 
diameter  than  the  threaded  part,  when  the 
bolt  is  under  strain  the  two  parts  will  not 
stretch  uniformly.     For  this  reason  the  bolt 
is   most  liable  to  break  where  the  threaded 

part  joins  the  other  part.     To  overcome  this  a  pIG    159 

hole  is  sometimes  drilled  from  the  center  of  the 

head  to  the  beginning  of  the  threaded  part.     This  hole  is  made  of  such 

size  that  the  cross-sectional  area  of  the  body  is  the  same  as  that  at  the 

root  of  the  thread. 

Find  the  diameter  of  the  hole  to  be  drilled  in  the  following  bolts  in 
accordance  with  the  preceding: 

(a)  Diameter  of  bolt  f  in.  with  10  U.  S.  S.  threads  to  1  in. 

(b)  Diameter  of  bolt  1|  in.  with  5  U.  S.  S.  threads  to  1  in. 

Ans.   (a)  0.422-  in.;  (b)  0.952  in. 

65.  In  computing  the  safe  working  pressure  for  a  steam  boiler,  a  factor 
of  safety  of  5  is  used.     That  is,  the  safe  working  pressure  is  1  of  the  burst- 
ing pressure.     The  bursting  pressure  in  pounds  per  square  inch  is  given 
by  the  formula 


where  P=  bursting  pressure  in  pounds  per  square  inch, 

T—  tensile  strength  of  boiler  plate  per  square  inch, 
I  =  thickness  of  boiler  plate  in  inches, 
?•  =  radius  of  boiler  in  inches, 
and       ft  =  a  constant  depending  upon  the  riveting  and  is  0.56  for  single 


212 


I'RA  CTICAL  MA  THEM  A  TICS 


riveted  boilers,  0.70  for  double  riveted  boilers,  and  0.88  for  triple  riveted 
boilers.     Derive  the  formula.     (See  Fig.  160.) 

66.  Find  the  bursting  and  the  safe  working  pressure  for  a  double 
riveted  boiler  66  in.  in  diameter,  made  of  plate  -fs  in.  thick,  if  tensile 
strength  is  50,000  Ib.  per  square  inch.  Am.  331.4  lb.;  66.3  Ib. 

67.  What  would  be  the  bursting  pressure  in 
pounds  per  square  inch  of  a  wrought  iron  pipe 
having  an  inside  diameter  of  3  in.  and  a  shell 
i  in.  thick  ?     Use  a  tensile  strength  of  40,000  lb. 
per  square  inch.  Ans.  3333  J  lb. 

68.  Holes  are  punched  in  sheets  of  metal  by 
means  of  great  pressure  applied  by  a  punch 
press.     The    pressure   is  usually   reckoned   at 
60,000  lb.  per  square  inch  of  surface  cut  over. 
For   example,    a   hole   2   in.   in  circumference 
punched  in  a  i-in.  plate  would  require  a  pres- 
sure of  2Xi  X60.000  lb.,  that  is,  the  area  of  the 
cylindrical  surface  sheared  off  times  60,000  lb. 

Find  the  pressure  necessary  to  punch  a  hole,  having  a  diameter  of  i  in., 
through  a  steel  plate  i  in.  thick.  Ans.  11,781  lb. 

69.  Find  the  pressure  necessary  to  punch  at  one  blow  a  hole  f  in.  in 
diameter  and  a  rectangular  hole  J  in.  by  J  in.  through  a  steel  plate  $  in. 
thick.  .4ns.  130,686  lb. 


FIG.    1(51. 

70.  Find  the  blow  necessary  to  cutout  the  corner  squares  and  the  holes 
in  the  box  shown  in  Fig.  161.     The  dimensions  are  as  given  and  the 
thickness  of  the  sheet  steel  is  &  in.  Ans.  17,678  lb. 

71.  Many  small  metal  articles  in  common  use  are  punched  out  of 
sheet  metal  and  pressed  into  shape.     The  blank  is  usually  cut  so  as  to 
have  the  same  area  as  the  area  of  the  finished  article  (Fig.  162).     For 
example,  the  blank  for  a  cylindrical  box,  having  a  diameter  of  1  in.  and 
a  depth  of  2  in.,  would  have  an  area  equal  to  the  combined  area  of  the 
sides  and  bottom  of  the  box.     Find  the  area  and  diameter  of  the  blank 
for  this  box. 


CYLINDERS 


213 


Solution.     Area  =  3. 1416X(2)2+3.1416 X 1  X 2  =  7.06  sq.  in.  89 
Diameter  of  blank  =  V7.0686  ^0.7854  =3  in. 

Note.     In  shallow  articles,  as  pail  covers,  the  diameter  of  the  blank  is 
often  found  by  adding  twice  the  depth  to  the  diameter  of  the  top. 


FIG.   162. 

72.  Find  the  diameter  of  the  blank  for  a  pail  cover  whose  diameter  is 
8  in.  and  depth  f  in.      Work  by  both  methods  suggested  above  and  com- 
pare results. 

73.  An  aluminum  cap  for  a  paste  bottle  has  the  dimensions  given  in 
Fig.  163.     Find  the  diameter  of  the  blank  from  which  it  was  pressed. 

Ans.  2.035  in. 

74.  A  shoe-blacking  box  has  a  diameter  of  3  in.  and  a  depth  of  1  in. 
Find  the  diameter  of  the  blank  from  which  it  was  pressed. 

Ans.  4.58  in. 


CHAPTER  XVI 
PYRAMIDS,  CONES,  AND  FRUSTUMS 

168.  Pyramid. — A  pyramid  is  a  solid  whose  base  is  a 
polygon,  and  whose  sides  are  triangles  with  their  vertices  at  a 
common  point,  called  the  vertex  or  apex  of  the  pyramid.  A 
pyramid  is  triangular,  square,  hexagonal,  etc.,  according  as  its 
base  is  a  triangle,  square,  hexagon,  etc. 

A  right  pyramid,  or  a  regular  pyramid,  is  a  pyramid  whose 
base  is  a  regular  polygon,  and  the  sides  or  faces  equal  isosceles 
triangles. 

Fig.  164  is  a  regular  pyramid  with  a  square  base. 


Fio.   164. 

In  a  regular  pyramid  the  axis,  or  the  line  drawn  from  the  ver- 
tex to  the  center  of  the  base,  is  perpendicular  to  the  base. 
This  line  is  the  altitude  of  the  pyramid. 

In  Fig.  164,  OF  is  the  altitude. 

The  slant  height  of  a  right  pyramid  is  the  line  drawn  from 
the  vertex  to  the  center  of  one  edge  of  the  base. 

EF  of  Fig.  164  is  the  slant  height, 

A  lateral  edge  is  the  line  in  which  two  faces  meet. 

BF  of  Fig.  164  is  a  lateral  edge. 

169.  Cone. — A  circular  cone  is  a  solid  whose  base  is  a 
circle,  and  whose  lateral  surface  tapers  uniformly  to  a  point, 

214 


PYRAMIDS,  CONES,  AND  FRUSTUMS 


215 


called  the  vertex  or  apex.     The  axis  of  the  cone  is  a  straight 
line  drawn  from  the  vertex  to  the  center  of  the  base. 

A  right  circular  cone  is  a  cone  whose  base  is  a  circle  and 
whose  axis  is  perpendicular  to  the  base. 

In  Fig.  165,  F-ABC  is  a  right  circular  cone. 

This  might  also  be  defined  as  a  solid  formed  by  a  right 
triangle    revolved    about 
one  of  its  legs  as  an  axis. 
It  may   be  called  a  cone 
of  revolution. 

The  altitude  of  a  cone 
is  the  perpendicular  line 
from  the  vertex  to  the 
base.  The  slant  height 
is  a  straight  line  drawn 
from  the  vertex  to  the  cir- 
cumference of  the  base. 

In  Fig.  165,  OF  is  the  altitude,  and  CF  the  slant  height. 

170.  Frustum. — If  the  top  of  a  pyramid  or  a  cone  is  cut 
off  by  a  plane  parallel  to  the  base,  the  remaining  part  is  called 
a  frustum  of  a  pyramid  or  a  cone. 

In  Fig.  166,  (n)  and  (b)  are  frustums. 
H  G 


FIG.  166. 

The  altitude  of  a  frustum  is  the  perpendicular  between  the 
bases,  as  NM  of  Fig.  166.  The  slant  height  of  the  frustum 
of  a  right  pyramid  or  cone  is  the  shortest  line  between  the 
perimeters  of  the  two  bases.  It  is  perpendicular  to  the  edge 
of  each  base  in  the  frustum  of  a  right  pyramid;  and  is,  there- 
fore, the  altitude  of  the  trapezoids  that  form  the  faces  of 
the  frustum. 

In  Fig.  16tJ(.o),  1>Q  is  the  slant  height. 


216  PRACTICAL  MATHEMATICS 

171.  Areas. — The  lateral  area  of  a  right  pyramid  is  found 
by  taking  the  sum  of  the  areas  of  the  triangles  forming  the 
faces  of  the  pyramid.     Since  the  altitudes  of  these  triangles  are 
each  the  slant  height  of  the  pyramid,  they  are  equal.     Because 
the  base  of  a  right  pyramid  is  a  regular  polygon,  the  bases  of 
the  triangles  are  equal.     We  then  have  the  following: 

RULE.  The  lateral  area  of  a  right  pyramid  or  cone  equals  the 
perimeter  of  the  base  times  one-half  the  slant  height. 

The  total  area  equals  the  lateral  area  plus  the  area  of  the  base. 

Since  the  faces  of  the  frustum  of  a  pyramid  are  trapezoids, 
we  have,  by  use  of  formula  [8],  the  following: 

RULE.  The  lateral  area  of  the  frustum  of  a  right  pyramid  or 
cone  equals  one-half  the  sum  of  the  perimeters  of  the  two  bases 
times  the  slant  height. 

The  total  area  equals  the  lateral  area  plus  the  areas  of  the  two 
bases. 

That  these  rules  apply  to  the  cone  as  well  as  to  the  pyramid 
may  be  seen  by  thinking  of  the  cone  as  a  pyramid  with  a  very 
great  number  of  sides  to  the  base. 

Using  S  for  lateral  area,  T  for  total  area,  h  for  altitude,  5 
for  slant  height,  p  for  perimeter  (P  and  p  for  frustum),  A 
for  area  of  base  (B  and  b  for  frustum),  the  rules  may  be  written 
as  the  formulas: 

[47]  S  =  £ps,  for  pyramid  or  cone. 
[48]  T  =  ^ps+A,  for  pyramid  or  cone. 
[49]  S  =  i(P+p)s,  for  frustum. 
[50]  T=i(P+p)s+B+b,  for  frustum. 

172.  Volumes. — A   particular   case   of   the    volume    of    a 

pyramid  is  seen  as  follows:  The  cube  of 
Fig.  167  is  divided  into  six  equal  pyra- 
mids with  their  vertices  at  the  center  of 
the  cube.  The  volume  of  the  cube 
equals  the  &rc&ABCD  times  the  altitude 
PQ.  Now  the  volume  of  one  of  the  six 
pyramids,  as  0-ABCD,  equals  $  of  the 
Flo  !  07  volume  of  the  cube,  and  hence  equals  the 

area  of  the  base  A  BCD  times  £  of  PO. 

RULE.  The  volume  of  a  pyramid  or  a  cone  equals  the  area  of 
the  base  times  one-third  the  altitude. 


PYRAMIDS,  CONES,  AND  FRUSTUMS  217 

Which  may  be  written  as  the  formula: 
[51]  V  =  iAh. 

The  volume  of  the  frustum  of  a  pyramid  or  cone  is  best 
stated  in  the  following  formula  : 

[52]  y  =  J 


The  volume  of  a  frustum  of  a  cone  is  usually  more  easily 
found  by 

[53]  V  =  i7rh(R2+r2+Rr), 

or  [54]  V  =  -iVh(D2+d2+Dd), 

where  R  and  D  are  radius  and  diameter 
respectively  of  lower  base,  and  r  and  d  of 
upper  base. 

It  should  be  noted  that  the  volume  of  a 
pyramid  is  one-third  the  volume  of  a  prism 
of  the  same  base  and  altitude,  and  that  the  FJG  1G8 

cone  bears  a  like  relation  to  the  cylinder. 

Example.     Find  the  volume  and  the  lateral  area  of  a  right 
cone  of  diameter  16  in.  and  altitude  12  in. 
Solution.     A  =  7rr2  =  3.  1416  X82  =  201.0624. 

V=iAh  =  3X201.0624X12  =  804.25  in.3 
s  =  \/122+82  =  14.422,    since   the   altitude,   ra- 
dius, and  slant  height  form  a  right  triangle  AOP  of  Fig.  168. 
S  =  ips  =  TITS  =  3.1416X8X14.422  =  362.465  in.2 

EXERCISES  50 

1.  Find  the  volume  of  a  right  cone  whose  altitude  is  8  in.  and  radius  of 
base  is  4.887  in.  Ans.  200  in.3  nearly. 

2.  Find  the  volume  and  total  area  of  a  cone  whose  radius  of  base  is 
6  in.  and  altitude  5.3  in.  Ans.  199.8+  in.3;  263.9  in.2 

3.  Find  the  volume  and  lateral  area  of  a  cone  whose  altitude  is  8  in. 
and  radius  6  in.  Ans.  301.6-  in.3;  188.5-  in.2 

4.  The  circumference  of  the  base  of  a  conical  church  steeple  is  35  ft. 
and  the  altitude  is  73  ft.     Find  the  lateral  area.  Ans.  1281  -  ft.2 

Suggestion.  Radius  of  base  =  5(35-^3.  1416).  The  altitude  and  radius 
are  the  altitude  and  base  of  a  right  triangle  of  which  the  hypotenuse 
is  the  slant  height  of  the  steeple. 

5.  Find  the  weight  of  a  conical  casting  of  iron  8  in.  in  diameter  and 
slant  height  14  in.  Ans.  58.4  Ib. 


'.'IS 


PR  A  CTICAI,  MA  THEM  A  TICS 


6.  A  ji.-iil  is  10  in.  in  diameter  on  the  bottom  and  12  in.  on  top.     If  tho 
slant  height  is  11  in.,  what  is  the  number  of  square  inches  of  tin  in  the 
pail?  Ann.  458.67. 

7.  Find  the  total  area  and  volume  of  a  cone  of  revolution  whose  alti- 
tude is  12  ft.,  and  the  diameter  of  whose  base  is  10  ft. 

Arm.  282.74  ft.  »;  314.16ft.' 

8.  The  diameter  of  the  top  of  a  water  pail  is  12  in.,  the  bottom  is  10  in., 
and  the  altitude  is  10  J  in.     How  many  quarts  will  the  pail  hold? 

Solution.    By[64],F  =  i»sX3.1416Xl0.5(121-|-101  +  12XlO)=1000.6in.:i 


1000.6  in.3  -=-57.75  in.3  =  17.33-  =  number  of  quarts. 

9.  Find  the  lateral  edge,  lateral  area,  and  volume  of  a  regular  pyramid, 
each  side  of  whose  triangular  base  is  10  ft.,  and  whose  altitude  is  18  ft. 
Ans.  18.90+  ft.;  273.45-  ft.1;  259.8+  ft.3 

10.  A  cone  12  in.  in  altitude  and  with  cir- 
cular base  8  in.  in  diameter  has  a  hole  2  in. 
in  diameter  bored  through  the  center  from 
apex  to  base.     Find  the  volume  of  the  part 
remaining.  Ans.   169.646  in.1 

Suggestion.  The  part  cut  away,  as  shown 
in  Fig.  169,  consists  of  a  cylinder  9  in.  in 
altitude  and  a  cone  3  in.  in  altitude.  The 
height  of  the  small  cone  can  be  found  from 
the  similar  triangles  AOP  and  BO'P  in 
which  we  have  the  proportion 

AO:B(y=OP:  O'P, 
or  4:  1  =  12:  O'P.     .:  O'P  =3. 

11.  Find  the  weight  of  a  green  fir  log  215 
ft.  long,  4  ft.  6  in.  in  diameter  at  one  end, 

and  20  in.  in  diameter  at  the  other  end,  the  specific  gravity  of  fir  being 
0.78.  Ans.  42  tons  nearly. 

12.  Hard  coal  dumped  in  a  pile  lies  at  an  angle  of  30°  with  the  hori- 
zontal.    Estimate  the  number  of  tons  in  a  pile  of  conical  shape  and  10  ft. 
high.     Large  egg  size  weighs  38  Ib.  per  cubic  foot.     (See  Fig.  170.) 

Ans.  60  tons  nearly. 
Suggestion.     Radius  of  base  of  pile  =  10X-s/3  by  Art.  112. 

13.  Find  the  number  of  tons  of  large  egg  coal 
in  a  pile  20  ft.  broad  and  100  ft.  long  with  cir- 
cular ends.  Ans.   99  tons  nearly. 

14.  A  tank  of  reen  forced  concrete  is  160  ft. 
long,  100  ft.  wide,  and  10ft.  6  in.  deep  outside 
dimensions.     The  side  walls  arc  8  in.  thick  at 
the  top   and    18  in.  at  the  bottom,  with  the 
slope  on  the  inside.     The  bottom  is  6  in.  thick. 

Find  the  number  of  cubic  yards  of  cement  in  the  tank  and  the  capacity 
of  the  tank  in  barrels.  Ans.  503  +  ;  36,670  bbl.  nearly. 


FIG.  170. 


PYRAMIDS,  CONES,  AND  FRUSTUMS  219 

15.  Find  the  weight  of  a  tapered  brick  stack  of  10  ft.  inside  diameter, 
with  a  wall  4  ft.  thick  at  the  base,  1  ft.  6  in.  at  the  top,  and  175  ft.  high. 
A  cubic  foot  of  brick  weighs  112  Ib.  Ans.  1095.5  tons. 

Suggestion.  Find  the  volume  of  a  frustum  of  a  cone  with  lower  base 
18  ft.  in  diameter  and  upper  base  13  ft.  in  diameter.  Subtract  from  this 
the  volume  of  the  cylinder  10  ft.  in  diameter. 

16.  A  cast-iron  driver  in  the  form  of  a  frustum  of  a  square  pyramid  is 
used  on  a  pile-driving  machine.     Find  the  weight  of  the  driver  if  it  is 
16  in.  high,  10  in.  square  at  the  bottom,  and  7  in.  square  at  the  top. 

Ans.  303.7  Ib. 

17.  A  cast-iron  cone  pulley  is  34  in.  long.     The  diameter  of  one  end  is 
12  in.  and  of  the  other  end  is  5  in.     A  circular  hole  2  in.  in  diameter 
extends  the  length  of  the  pulley.     Find  the  weight  of  the  pulley. 

Ans.  502.2  Ib.  • 


CHAPTER  XVII 
THE  SPHERE 

173.  Definitions. — A  sphere  is  a  solid  bounded  by  a  curved 
surface,  every  point  of  which  is  equally  distant  from  a  point 
within,  called  the  center.     A  straight  line  passing  through  the 
center  and  ending  in  the  surface  is  called  a  diameter.     A 
line  extending  from  the  center  to  the  surface  is  a  radius. 

If  the  sphere  is  cut  by  a  plane,  the  section  is  a  circle.  If  the 
section  is  through  the  center  of  the  sphere,  it  is  called  a  great 
circle;  if  not  through  the  center,  it  is  called  a  small  circle. 

The  circumference  of  a  sphere  is 
the  same  as  the  circumference  of  a 
great  circle. 

In  Fig.  171,  circles  ACB  and  NCS  arc 
great  circles,  and  MER  is  a  small  circle. 

The  parallels  of  latitude  on  the 
surface  of  the  earth  are  small  circles. 
The  meridians  of  the  earth  all  run 
through  both  the  north  and  the 
south  poles  and,  therefore,  are  great 
circles. 

174.  Area. — The  following  is  proved  in  geometry: 

RULE.     The  area  of  the  surface  of  a  sphere  equals  four  times 
the  area  of  a  circle  of  the  same  radius. 
Or  stated  as  a  formula: 

[55]  S  =  47rr2  =  7rd2, 

where  »S  is  the  area  of  the  surface  of  the  sphere,  r  the  radius,  and 
d  the  diameter. 

The  student  may  satisfy  himself  that  this  is  true  by  winding 
evenly  the  surface  of  a  ball  with  heavy  cord,  and  then  coiling 
the  same  cord  into  four  circles  of  the  same  radius  as  the  radius 
of  the  sphere. 

The  rule  can  be  derived  from  the  fact  that  a  sphere  has  the 

220 


THE  SPHERE 


221 


same  area  as  the  lateral  area  of  a  cylinder  having  the  same 
radius  as  the  sphere,  and  an  altitude  equal  to  the  diameter  of 
the  sphere.     Area  of  lateral  surface  of  cylinder  =  2xr  X  2r  =  4irr2. 
175.  Volume. — Geometry  gives  the  following: 
RULE.     The  volume  of  a  sphere  equals  the  area  of  the  surface 
times  one-third  of  the  radius. 


(a) 


FIG.   173. 


FIG.  172. 
Or  stated  as  a  formula: 

The  reasonableness  of  this  may  be  seen  by  thinking  of  the 
surface  of  the  sphere  as  divided  into  a  large  number  of  small 
polygons.  Let  these  be  so  small  that  they  may  be  considered 
as  planes.  Now  if  we  think  of  the 
sphere  cut  into  pyramids  having 
these  polygons  as  bases,  and  having 
their  vertices  at  the  center  of  the 
sphere,  as  shown  in  Fig.  173,  the 
volume  of  one  of  these  small  pyra- 
mids, represented  in  (a)  of  the  figure, 
is  found  by  [51]  to  be  fr  times  the 
area  of  the  small  polygon.  And 
the  volume  of  all  the  small  pyramids 
is  equal  to  the  whole  surface  of  the 
sphere  times  |r.  Hence  V  —  %Sr. 

176.  Zone  and  segment  of  sphere. — A  portion  of  the  volume 
of  a  sphere  included  between  two  parallel  planes  is  a  segment 
of  the  sphere.  If  both  the  planes  cut  the  surface  of  the  sphere, 
the  segment  is  a  segment  of  two  bases.  In  Fig.  174,  the 
segment  between  the  planes  ABC  and  DEF  is  a  segment  of 


222  I'RACTICAL  MATHEMATICS 

two  bases.     The  part  of  the  sphere  above  DBF  is  a  segment 
of  one  base. 

That  portion  of  the  surface  of  the  sphere  between  two 
parallel  planes  is  a  zone.  The  altitude  of  the  segment  or 
zone  is  the  perpendicular  between  the  parallel  planes. 

Thus,  OQ  is  the  altitude  of  the  segment  between  the  planes  ABC 
and  DBF. 

Here  we  can  neither  derive  the  rules  for  the  area  of  a  zone 
and  the  volume  of  a  segment  nor  make  them  seem  reasonable 
by  any  discussion.  They  are  of  some  importance  practically, 
especially  the  volume  of  the  segment. 

RULE.  The  area  of  a  zone  is  equal  to  the  circumference 
of  a  great  circle  of  the  sphere  times  the  altitude  of  the  zone. 


Segment  of  One  Ease  Segment  of  Two  Bases 

Fiu.    175. 

This  rule  is  stated  in  the  formula: 

[67]  Z  =  27rrh, 

where  Z  is  the  area  of  the  zone,  h  the  altitude,  and  r  the 
radius  of  the  sphere. 

It  is  readily  seen  from  formula  [57]  that  the  area  of  any 
two  zones  on  the  same  or  equal  spheres  are  to  each  other  as 
their  altitudes.  It  also  follows  that  any  zone  is  to  the  sur- 
face of  the  sphere  as  the  altitude  of  the  zone  is  to  the  diam- 
eter of  the  sphere. 

If  a  sphere  is  cut  by  parallel  planes  that  are  equal  distances 
apart,  as  the  planes  cutting  the  sphere  in  Fig.  174,  then  the 
zones  are  all  equal. 

Since  the  parallels  of  30°  north  and  south  latitude  are  in 
planes  that  bisect  the  radii  drawn  to  the  north  and  south 
poles,  then  one-half  of  the  surface  of  the  earth  is  within  30° 
of  (he  equator. 


THE  SPHERE  223 

The  volume  of  a  spherical  segment  is  given  by  the  formula: 

[58]  V  =  ih7r(r12+r22)  +  i7rh3, 

where  V  is  the  volume,  h  the  altitude,  and  n  and  r2  the  radii 
of  the  bases  of  the  segment.  If  the  segment  has  only  one 
base,  one  of  the  radii  is  zero. 

Example  1.     Find  the  surface,  volume,  and  weight  of  a 
cast-iron  ball  of  radius  12|  in. 

Solution.     By  [55],  S  =  4*r*  =  4 X 3.1416 X12.52  =  1963.5  in.2 
By  [56],  F  =  |Sr= 1X1963.5X12.5  =  8181. 25  in.3 
Since  1  in.3  of  cast  iron  weighs  0.26  lb., 

the  weight  =  0.26  Ib.X  8181. 25  =  2127. 125  lb. 
Example  2.  A  sphere  8  in.  in 
radius  is  cut  by  two  parallel  planes, 
one  passing  2  in.  from  the  center, 
and  the  other  6  in.  from  the  center. 
Find  the  area  of  the  zone,  and  the 
volume  of  the  segment  between  the 
two  planes,  if  both  planes  are  on  the 
same  side  of  the  center. 

Solution.     By  [57],  Z  =  2irrh  =  2X3.1416X8X4  =  201. 06  in.2 
ri  =  \/(OB}2-(OW=  VS^T2  =  A/60; 
r2  =  A/(OZ))2-  (OF)2=  A/~82-62=  A/28. 
By  [58],  V^hirW+rtf  +  brh* 

=  |X4X3. 1416(60  +  28) +  |X3. 1416  X43  =  586.43  in.3 

EXERCISES  51 

1.  Find  the  volume  and  the  area  of  the  surface  of  a  sphere  6  ft.  in 
diameter.  Ans.   113.1  -  ft.3;  113.1- ft.2 

2.  A  sphere  4  in.  in  radius  is  cut  from  a  cylinder  8  in.  high  and  8  in. 
in  diameter.     Find  the  volume  cut  away.  Ans.  134.04+  in.3 

3.  The  radius  of  a  sphere  is  2  ft.     Find  the  area  of  the  surface  and  the 
volume.  Ans.  50.266 -ft.2;    33.51+ ft.3 

4.  How  much  will  a  sphere  of  cast  iron  weigh  if  it  is  3  in.  in  diameter, 
and  if  cast  iron  weighs  0.26  lb.  per  cubic  inch?  Ans.  3.675  lb. 

5.  A  cubic  foot  of  lead  weighs  712  lb.     Find  the  weight  of  a  ball  3  in. 
in  diameter.  Ans.  5.825  lb. 

6.  How  many  square  feet  of  tin  will  it  take  to  roof  a  hemispherical 
dome  40  ft.  in  diameter?  Ans.  2513+  ft.2 

7.  Find  how  many  acres  of  land  on  the  surface  of  the  earth,  if  one- 
fourth  of  the  surface  is  land  and  the  radius  is  4000  miles. 

Ans.  32,170,000,000  nearly. 


PRACTICAL  MA  THEM  A  TICS 


8.  Find  the  volume  of  a  cylinder  2  ft.  in  diameter  and  2  ft.  in  altitude; 
of  a  sphere  2  ft.  in  diameter;  and  of  a  cone  2  ft.  in  diameter  and  2  ft.  in 
altitude.     Compare  the  three  volumes.     (See  Fig.  177.) 

9.  Find  the  volume  of  the  segment  between  two  parallel  planes  6  in. 
apart  that  cut  a  sphere  12  in.  in  radius,  if  one  plane  passes  2  in.  from  the 
center.     There  are  two  cases:  (a)  when  the  center  of  the  sphere  lies  out- 
side of  the  segment,  and  (b)  when  the  center  lies  in  the  segment. 

Ans.  (a)  2186.6- in.*;  (b)  2638.9+ in.1 

10.  An  iron  ball  3  in.  in  diameter  has  a  coating  of  lead  1  in.  thick. 
Find  the  volume  of  the  iron,  of  the  lead,  and  the  weight  of  each. 

Ans.  14.137  in.8;  51.313  in.»;3.6T6-  lb.;21.141-  Ib. 

11.  A  ball  of  lead  2  in. 
in     diameter    is    pounded 
into   a  circular  sheet  0.01 
in.  thick.     How  large  in  di- 
ameter is  the  sheet? 

Ans.  23  in.  nearly. 

12.  A  water  tank,  6  ft. 
Fio.  177.                                 in  total  length  and   18  in. 

in  diameter,  is  in  the  form 

of  a  circular  cylinder  with  two  hemispherical  ends.  Find  its  capacity 
in  gallons.  Ans.  72.7+  gal. 

13.  A  hollow  copper  sphere  used  as  a  float  weighs  10  oz.  and  is  5  in.  in 
diameter.     How  heavy  a  weight  will  it  support  in  water? 

Ans.  Less  than  27.9  oz. 

Suggestion.  It  will  support  a  weight  less  than  the  weight  of  water 
displaced  by  the  sphere  minus  the  weight  of  the  sphere. 

14.  A  circular  flower  bed  in  a  park  is  25  ft.  in  diameter  and  is  raised 
2  ft.  6  in.  in  the  center  making  a  spherical  segment.     How  many  loads  of 
dirt  did  it  take  to  build  it  up  if  one  load  is  1 J  cu.  yd.?     Ans.  15  J  nearly. 

15.  In  a  practical  hand- 
book  the  following   rule   is 
given  as  nearly  correct.      In 
fact,  it  is  correct.     The  area 
of   a  flanged  spherical  seg- 
ment, a  vertical  section  of 
which  is  shown  in  Fig.  178, 
is    equal   to   the   area   of  a 

circle  of  radius  equal  in  length  to  the  line  drawn  from  the  top  of  the  seg- 
ment to  the  edge  of  the  flange,  that  is,  equal  to  a  circle  of  radius  AB. 

Find  the  area  of  a  flanged  segment  having  dimensions  as  given  in  Fig. 
178.  Work  both  by  the  rule  and  by  using  the  formulas  for  area  of  a  ring 
and  of  :i  zone. 

Solution.      By  [16],   radius,   r,   of  sphere  of  which  zone  is  a  part 


FIG.   178. 


5* +2* 
!  2X2 


=  7.25. 


THE  SPHERE 


225 


By  [57],  7  =  2X3.1416X7.25X2=91.1064  in.2 

By  [28],  area,  A,  of  ring  =  3.1416(8+5)(8-5)  =122.5224  in.3 

91.1064  in.2  +  122.5224  in.2  =213.63-  in.2 

=  V68. 


FIG.  179. 


AB=- 

Area  of  circle  having  a  radius  =V<38  is  3. 1416  X  (\/68)2  =  213.63  -in.2, 
which  is  the  same  as  the  result  by  the  first  method. 

16.  Find  the  per  cent  of  error  in  using  the  following  rule:  To  find 
the  weight  of  a  cast-i/on  ball  mul-  ()  „ 
tiply  the  cube  of  the  diameter  in 

inches  by  0.1377,  and  the  product 
is  the  weight  in  pounds. 

Ans.  Rule  is  correct  if  1  cu.  in.  of 
cast  iron  weighs  0.263  Ib. 

17.  Fig.   179  is  the  vertical  cross 
section  of  a  casting,  the  inner  and 

outer  "skins"  being  spherical  zones.     Find  the  weight  of  metal  at  0.35 
Ib.  per  cubic  inch  necessary  to  make  the  casting. 

Ans.  176  Ib.  nearly. 
Suggestion.      Find  the  difference  of  the  volumes  of  the  two  segments. 

18.  A  hemispherical  cap  of  aluminum  is  3 5  in.  in  diameter.     Find  the 
diameter  of  the  blank  from  which  it  is  pressed.  Ans.  4.95  —  in. 

19.  Find  the  diameter  of  the  blank  if  the  cap  in  the  preceding  exercise 
has  a  flat  ring  ?  in.  wide  around  it.  Ans.  5.70+   in. 

20.  Show  that  the  volume  of  a  round  or  button  head  of  a  machine  screw 
is  given  by  the  formula 

V=irh 


where  D  is  the  diameter  of  the  head  and  h  the 
height  of  the  head. 

Suggestion.     In  formula  [68], 


But  n  =—  and  r2  =0. 


21.  Find  the  volume  of  the  head  of  a  round 
head  machine  screw  if  the  diameter  of  the  head 
is  0.731  in.  and  the  height  is  0.279  in. 

Ans.  0.070  cu.  in. 

22.  The  water  tank  shown  in  Fig.  180  con- 
sists  of   a   cylinder  with  a  hemisphere  below. 

The  diameter  is  20  ft,  and  the  height  of  the  cylindrical  part  is  22  ft. 
Find  the  capacity  of  the  tank  in  gallons.  Ans.  67,369  gal. 


CHAPTER  XVIII 
VARIOUS  OTHER  SOLIDS 

177.  Anchor  ring. — A  ring  formed  of  a  cylinder  bent  into 
a  circular  form,  as  in  Fig.  181,  is  called  an  anchor  ring.  The 
mean  length  of  the  rod  in  such  a  ring  is  the  circumference  of  a 
circle  of  radius  ON. 

Any  cross  section  of  such  a  ring  will  be  a  circle.  Since  the 
ring  may  be  considered  as  a  cylinder  bent  into  circular  form, 
the  area  of  the  surface  is  2-irXONX circumference  of  a  cross 
section.  If  ON  =  R,  and  the  radius  of  the  cross  section  NM  is 
r,  we  have  for  the  area  the  formula: 

[59]  A  =  27rRX27rr  =  47T2Rr. 

The  volume  is  the  same  as 
the  volume  of  a  cylinder  with  an 
altitude  that  is  equal  to  the 
mean  circumference  of  the  ring, 
hence  the  following: 

[60]  V  =  27rRX7rr2  =  27r2Rr2. 

These  rules  may  be  generalized 
so  as  to  apply  to  any  circular  ring. 
In  general,  the  area  of  the  sur- 
face equals  the  perimeter  of  the  cross  section  times  the  cir- 
cumference drawn  through  the  center  of  gravity  of  the  cross 
section.  The  volume  equals  the  area  of  the  cross  section 
times  the  circumference  drawn  through  the  center  of  gravity 
of  the  cross  section. 

EXERCISES  62 

1.  The  cross  section  of  a  solid  wrought-iron  ring  is  a  circle  of  4  in. 
radius.     The  inner  radius  of  the  ring  is  3  ft.     Find  the  area  of  the  surface 
and  the  volume  of  the  ring.  Ans.  6316.6  in.';  12,633.2  in.3 

2.  The  cross  section  of  the  rim  of  a  flywheel  is  a  rectangle  6  in.  by 
8  in.,  the  shorter  dimension  bring  in  the  diameter  of  the  wheel.     The 

220 


VARIOUS  OTHER  SOLIDS 


227 


wheel  is  22  ft.  in  outer  diameter.     Find  the  volume  of  the  rim  and  its 
weight  if  of  cast  iron.  Ans.  22.515  ft.3;  10,132  Ib. 

3.  Find  the  weight  of  a  cast-iron  water  main  12  ft.  in  length,  2  ft.  in 
outer  diameter,  and  1  in.  thick.     Solve  both  by  considering  it  as  a  ring 
and  as  a  hollow  cylinder.  Ans.  2709.6  Ib. 

4.  Find  the  area  of  the  surface  and  volume  of  a  ring  of  outer  diameter 
10  in.,   made  of  round  iron   1  in.  in  diameter.     What  is  its  weight  at 
0.28  Ib.  per  cubic  inch?  Ans.  88.83  in.2;  22.207  in.3;  6.218  Ib. 

6.  An  anchor  ring,  13  in.  in  outer  diameter,  of  1J  in.  round  iron,  has 
the  same  volume  as  what  length  of  a  bar  1£  in.  by  If  in.  in  cross  section? 

Ans.  24.16  in. 

6.  Find  the  weight  of  an  anchor  ring  of  cast  iron,  outer  diameter  3  ft., 
the  iron  being  circular  in  cross  section  and  6  in.  in  diameter.  (Use 
450  Ib.  per  cubic  foot.)  Ans.  693.9  Ib. 


FIG.   182. 

178.  Prismatoids. — A  prismatoid  is  a  solid  whose  bases  are 
parallel  polygons  and  whose  faces  are  quadrilaterals  or  trian- 
gles. One  base  may  be  a  point,  in  which  case  the  prismatoid 
is  a  pyramid;  or  one  base  may  be  a  line,  in  which  case  it  is 
wedge-shaped.  The  rule  for  finding  the  volumes  of  prismatoids 
holds  in  very  many  cases  when  the  faces  become  curved  sur- 
faces and  the  prismatoid  has  become  a  cone,  frustum  of  a 
cone,  a  cylinder,  a  sphere,  a  spindle  of  some  kind,  or  one  of 
various  other  forms  that  cannot  well  be  described  here.  The 
rule  is  as  follows: 

RULE.  To  find  the  volume  of  a  prismatoid,  add  together 
the  areas  of  the  two  bases  and  four  times  the  area  of  a  section 
midway  between  them  and  parallel  to  them,  then  multiply  the 
sum  by  one-sixth  the  perpendicular  between  the  bases. 


PR  A  CTICA  L  MA  Til  EM  A  TICK 


The  rule  may  he  stated  in  the  formula: 

[61J  V=Jh(B1+4M+B2), 
where  B\  and  #2  are  the  two  bases  and  M  the  mid-section. 

In  Fig.  182  are  given  some  forms  to  which  the  rule  for  the 
volume  of  a  prismatoid  will  apply.  The  dimensions  of  the 
mid-section  may  be  found  by  actually  measuring  the  lines  or 
by  computing  them. 


Example.  By  formula  [61],  find  the  volume  of  a  frustum 
of  a  pyramid  in  which  the  bases  are  regular  hexagons  10  in. 
and  6  in.  on  a  side  respectively  and  whose  altitude  is  18  in. 
(See  Fig.  183.) 

h— "->! 


Fro.  185.  FIQ.  186. 

Solution.     AB  =  $(1Q  in. +  6  in.)  =8  in. 
Area  of  lower  base  =52X1.732X6  =  #,. 
Area  of  upper  base  =  32  X  1.732  X  6  =  B2. 
4Xarea  of  mid-section  =4X42X1.732X6=43f. 
.'.  #i+4M +£2=  1018.416  in.2 
And  F  =  JX  18X1018.416  =  3055.248  in.3 

EXERCISES  63 

1.  Use  the  rule  for  volume  of  prismatoids,  and  find  the  volumes  called 
for  in  exercises  3,  page  217;  exercises  8  and  9,  page  218. 

2.  Use  formula  [61]  to  find  the  volume  of  a  hemisphere  and  check  by 
the  ordinary  rule. 


VARIOUS  OTHER  SOLIDS 


229 


3.  Use  formula  [61]  to  find  the  volume  of  the  solid  shown  in  Fig.  184. 

Ans.  3420  in.3 

4.  Use  formula  [61]  to  find  the  volume  of  the  solid  shown  in  Fig.  185. 

Ans.  3360  in.3 


6.  A  concrete  pier  for  a  railway  bridge  has  the  dimensions  shown  in 
Fig.  186,  the  bases  being  rectangles  with  semicircles.  Find  the  number 
of  cubic  yards  of  concrete  in  21  such  piers.  Ans.  793.3+. 

6.  A  railroad  cut  has  the  dimensions  shown  in  Fig.  187,  which  shows 
the  vertical  section  and  three  cross  sections.  Find  the  volume  of  the 
earth  removed  in  cubic  yards.  Ans.  7972 1  yd.3 


PART  THREE 
ALGEBRA 


CHAPTER  XIX 

NOTATION  AND  DEFINITIONS 

179.  General  remarks. — In  mathematics,  the  attempt 
is  made  to  do  certain  things  more  easily  and  in  less  time  than 
they  can  otherwise  be  done.  In  arithmetic,  many  processes 
were  learned  that  saved  time  and  labor.  In  performing  these 
processes,  certain  signs  and  symbols  were  used  to  express  the 
ideas.  New  signs  and  symbols  were  introduced  as  they  were 
needed  to  express  the  new  ideas  that  were  involved. 

Thus,  there  were  used  the  numerals  0,  1,  2,  3,  4,  5,  6,  7,  8,  and  9;  the 
letters  of  the  alphabet;  and  various  signs  among  which  are  +,  — ,  X, 
-^,  (  ),  and  V- 

Algebra  is  a  name  applied  to  the  continuation  of  arithmetic. 
The  same  signs  and  symbols  are  used  in  algebra  as  in  arith- 
metic, and  they  have  exactly  the  same  meanings. 

From  time  to  time  as  we  proceed,  we  shall  find  it  convenient 
to  add  to  the  symbols  and  signs,  in  order  that  we  may  express 
new  ideas  or  perform  new  processes.  We  shall  also  expect  that 
many  new,  simpler,  and  more  powerful  methods  of  procedure 
will  be  developed;  in  fact,  this  is  the  chief  aim  in  continuing 
the  study  of  mathematics. 

189.  Definite  numbers. — The  numerals  0,  1,  2,  3,  etc., 
have  definite  meanings.  For  instance,  the  symbol  4  represents 
the  idea  we  call  four.  It  may  be  4  yards,  4  dollars,  4  pounds, 
or  4  of  any  other  units;  but,  in  any  case,  is  a  definite  number. 
We  have  learned  that  the  letter  TT  represents  a  definite  number, 
the  ratio  of  the  circumference  to  the  diameter  of  a  circle. 
This  cannot  be  expressed  exactly  by  the  numerals  1,  2,  3,  etc.; 
but  has,  none  the  less,  a  fixed  value. 

181.  General  numbers. — We  have  used  the  letter  b  to 
represent  the  number  of  units  in  the  base  of  a  triangle.  Its 

331 


232  PRACTICAL  MATHEMATICS 

value  changed  for  different  triangles,  that  is,  it  represented 
in  a  general  way  the  length  of  the  base  of  a  triangle.  Its 
value  might  be  given  as  10  ft.,  6  in.,  or  as  any  number  of  any 
sized  units  of  length.  Likewise,  r  represents  the  radius  of  a 
circle;  but  when  it  occurs  in  the  formula  A=7ir2,  we  do  not 
think  of  a  particular  value  for  it.  Such  an  idea  as  we  rep- 
resent by  b  or  by  r  cannot  be  represented  by  the  numerals. 
The  idea  is  a  general  number-idea.  It  is  usually  represented 
by  a  letter  of  the  alphabet. 

In  a  particular  discussion,  the  letter  or  letters  used  stand 
for  the  same  value  throughout  the  discussion.  For  example, 
when  we  are  considering  a  particular  circle  the  letter  r  repre- 
sents a  definite  length  as  10  ft. 

182.  Signs. — The  signs  +,  — ,  X,  and  -5-  are  signs  of 
operation.  They  continue  to  have  the  same  meaning  as  in 
arithmetic.  As  we  have  already  seen,  the  sign  X  is  not  often 
expressed  where  a  multiplication  is  indicated  between  numbers 
expressed  by  letters.  The  symbol  (•)  may  be  used  instead; 
but  usually  no  sign  is  expressed.  Thus,  aXb  is  written  a-b 
or  simply  ab.  Similarly,  2axy  means  2  time  a  times  x  times  y. 

The  signs  of  grouping  are  the  parentheses  (  ),  the  brackets 
[  ],  the  braces  {  },  and  the  vinculum  -  — .  The  first  three 
are  placed  around  the  parts  grouped,  and  the  vinculum  is 
usually  placed  over  what  is  grouped.  They  all  indicate  the 
same  thing;  namely,  that  the  parts  enclosed  are  to  be  taken 
as  a  single  quantity. 

Thus,  12  — (10— 4)  indicates  that  4  is  to  be  subtracted  from  10  and 
then  the  remainder  is  to  be  taken  from  12.  Hence  12  — (10  — 4)  =6. 
Exactly  the  same  thing  is  indicated  by  12  — [10  — 4),  12—  {10  —  4},  and 
12-10~4. 

The  vinculum  is  most  frequently  used  with  the  radical 
sign.  Thus,  \/6425. 

7  —  4 

It  is  to  be  noted  that  in  the  form  0     -A  the  horizontal  lino 

3+4 

serves  as  a  vinculum  and  as  a  sign  of  division.  It  thus  per- 
forms three  duties:  first,  indicates  a  division;  second,  binds 
together  the  numbers  in  the  numerator;  and  third,  binds 
together  the  numbers  in  the  denominator. 

In  performing  the  operations  in  a  problem  containing  the 


NOTATION  AND  DEFINITIONS  233 

signs  of  grouping,  the  operations  within  the  grouping  signs 
must  be  considered  first. 

183.  Algebraic    expression. — An    algebraic    expression    is 
any  expression  that  represents  a  number  by  means  of  the 
signs  and  symbols  of  algebra. 

A  numerical  algebraic  expression  is  one  made  up  wholly 
of  numerals  and  signs.  A  literal  algebraic  expression  is  one 
that  contains  letters. 

Thus,  14  +  13  — (4+3)  and  3ab  —  4cd  are  algebraic  expressions;  the 
first  is  numerical  and  the  second  is  literal. 

The  value  of  an  algebraic  expression  is  the  number  it 
represents. 

184.  Coefficient. — If  we  have  such  an  expression  as  8abx, 
8,  a,  b,  and  x  are  factors  of  the  expression.     Any  one  of  these 
factors  or  the  product  of  any  two  or  more  of  them  is  called  the 
coefficient  of  the  remaining  part. 

Thus,  Sab  may  be  considered  the  coefficient  of  x,  or 
Sa  the  coefficient  of  bx',  but  usually,  by  the  coefficient, 
we  mean  the  numerical  part  only.  It  is  then  called  the 
numerical  coefficient.  If  no  numerical  part  is  expressed,  1 
is  understood.  Thus,  laxy  is  the  same  as  axy. 

185.  Power,   exponent. — If  all   the   factors  in   a  product 
are  equal  as  a-a-a-a,  the  product  of  the  factors  is  called  a 
power  of  one  of  them.     The  form  a-a-a-a  is  usually  written  a4. 
The  small  number  to  the  right  and  above  indicates  how  many 
times  a  is  taken  as  a  factor.     (See  Arts.  74  and  75.) 

In  the  above  power  a  is  called  the  base  and  4  the  exponent. 

The  exponent  of  a  power  is  a  number  written  at  the  right 
and  a  little  above  the  base.  When  it  is  a  positive  whole 
number  it  shows  how  many  times  the  base  is  to  be  taken  as  a 
factor. 

Thus,  c2  is  read  c  square  or  c  second  power,  and  indicates  that  c  is 
taken  twice  as  a  factor;  c3  is  read  c  cube  or  c  third  power,  and  indicates 
that  c  is  taken  three  times  as  a  factor;  c4  is  read  c  fourth  power,  and  indi- 
cates that  c  is  taken  four  times  as  a  factor;  cn  is  read  c  nth  power  or  c  ex- 
ponent n,  and  indicates  that  c  is  taken  n  times  as  a  factor. 

When  no  exponent  is  written  the  exponent  is  understood  to 
be  1. 

Thus,  a  is  the  same  as  a1. 


234  PRACTICAL  MATHEMATICS 

186.  A  term  in  an  algebraic  expression  is  a  part  of  the 
expression  not  separated  by  a  plus  or  a  minus  sign. 

Thus,  in  4ox+3c  —  d,  4oz,  3c,  and  d  are  terms. 

It  is  convenient  to  have  names  for  algebraic  expressions 
having  different  numbers  of  terms.  A  monomial  is  an  alge- 
braic expression  consisting  of  one  term.  A  binomial  consists 
of  two  terms;  and  a  trinomial  consists  of  three  terms.  Any 
algebraic  expression  of  two  or  more  terms  is  called  a  poly- 
nomial or  a  multinomial. 

Terms  that  are  exactly  the  same  or  differ  only  in  their 
coefficients,  are  called  like  terms  or  similar  terms.  Terms 
that  differ  otherwise  than  in  their  coefficients,  are  unlike, 
or  dissimilar  terms. 

Thus,  60  !z*,  —  7a'z1,  and  16ajx*  are  like  terms;  while  6oxl,  —  7afz*, 
and  16ar/2  are  unlike  terms. 

187.  Remarks. — The  object  of  the  exercises  of  this  chapter 
is  to  recall  the  meanings  and  uses  of  signs  and  symbols,  and 
to  fix  in  mind  the  new  ideas  that  have  been  given  here.     The 
doing  of  these  exercises  must  not  be  slighted  by  the  student, 
for  he  must  become  familiar  with  the  mathematical  way  of 
stating  ideas  in  order  that  he  may  be  prepared  for  the  work 
that  comes  later. 

EXERCISES  64 

Find  the  value  of  each  of  the  following: 

1.  3+6-2+4-1+7+2. 

2.  7+2X6-3+8X2-2. 

3.  8  +  16-5-2-4  +  14-5-7+21. 

4.  75  +  10+25X6-3X3-15. 

5.  150-5-6+4X25-75. 

6.  150-5-5X6  +  10X9  +  17. 

Name  the  monomials  in  the  following  list,  the  binomials,  the  trinomials. 
Which  of  them  are  polynomials? 

7.  m'+2t>.  8.  7o»6*c.  9.  4+9c+d. 

10.  ofe+ac+od.          11.  a»+4c+<i.  12.  17a«6*n. 

13.  a*+6'-cJ.  14.  Igt*.  15.  Jirr'. 

16.  i/urr*.  17.  5z'+3y.  18.  7yz-4**-2». 

19.  19zj/-3z-2.        20.  \x-\y-\z.  21.  9zV+3zyz'+3zV- 

22.  What  is  the  coefficient  of  t>  in  Ex.  7?     Of  a'b'c  in  Ex.  8?     Of  c  in 
Ex.  8?     Of  a6  in  Ex.  12?     Of  r  in  Ex.  16? 

23.  Name  the  numerical  coefficients  in  Exs.  9,  13,  and  20. 

24.  Name  the  exponents  in  Exs.  8,  11,  and  18. 


NOTATION  AND  DEFINITIONS  235 

26.  How  many  factors  are  there  in  Ex.  8?     In  Ex.  12? 
Write  the  following  in  algebraic  symbols: 

26.  The  result  of  adding  7  times  a  to  9  times  b. 

27.  The  result  of  subtracting  10  times  a  from  6  times  n. 

28.  The  product  of  3  times  a3  times  b  square. 

29.  The  product  of  7  times  a  fourth  power  times  n  cube  times  d. 

30.  The  product  of  the  sum  of  a  and  b,  times  the  difference  found  by 
subtracting  b  from  a.  Ans.   (a-\-b)(a  —  b). 

31.  The  square  of  the  sum  of  a,  b,  and  c.  Ans.   (a+b+c)z. 

32.  The  quotient  of  the  sum  of  x  and  y  divided  by  the  difference  when 
y  is  subtracted  from  x. 

33.  Express  the  product  of  m3  and  the  sum  of  the  two  fractions,  2 

f  2      31 

divided  by  a  and  3  divided  by  b.  Ans.  m3    -  +  j-  1  • 

(a      o  j 

34.  Express  the  product  of  the  sum  of  a,  26,  and  3c,  and  the  sum  of 
3<z,  46,  and  7c.  Ans.   (o+26+3c)(3a+4&+7c). 

36.  Express  the  square  of  the  sum  of  a  and  b,  plus  the  cube  of  the  sum 
of  m,  n,  and  p.  Ans.   (a+b)2  +  (m+n+p).3 

Translate  the  following  algebraic  expressions  into  English  : 
36.  12n+5m.  37.  7n-14m.  38.  x(m  —  ri). 

39.  p2x2z.  40.   (3c+4d)(a+b).         41. 


.  44.    (Ml'. 

m  —  n  (x  —  yr  (b      a] 

45.  9a-(16-6).        46.  (x+y)2+b\  47.   f-^r  -  -^—.  \  x*. 

[a+o      a  —  oj 

48.  Write  the  following  in  a  more  compact  form  by  using  exponents: 
(1)  4aaabbbb,  (2)  27xxyyyzzz. 

(3)   17abbcccdddd.  (4)   14aa  +  1666  +  17ccc. 

49.  Write  in  a  more  compact  form  by  using  exponents: 
(1)  2-2-2-3-3.  (2)  5-5-5-3-S-2. 

(3)  7-4-4-5-5-5.  (4)  S-3-3-3-4-4-5-5. 

60.  Express  each  of  the  following  as  a  single  number  without  an  ex- 
ponent: 

(1)   105.  (2)  4X107.  (3)  2.3  X108. 

(4)  8.62  X107.  (5)  4.326  X1011.  (6)  2.763  X1011. 

61.  Express  the  following  large  numbers  as  a  small  number  times  a 
power  of  10: 

(1)  7000000000.         (2)  23460000000.  (3)  2456000000000. 

(4)  427600000000.      (5)  5678000000000.         (6)  54237600000000. 

62.  Find  the  value  of  (1)  2342,  (2)  73-22-5,  (3)  24-53-72. 

53.  Separate  the  following  into  prime  factors  and  express  compactly 
by  using  exponents  : 

(1)  864.     (2)   1296.     (3)  78400.     (4)  94500.     (5)   152100. 

64.  If  o  =  4,  what  is  the  value  of  3a?  Of  7aV  Of  a2?  Of  a3?  Of 
4a2? 

55.   What  is  the  value  of  15a  if  a  is  4,  7,  2'i,  0,  3',,  83,  5|? 


PRACTICAL  MATHEMATICS 

56.  If  I  is  the  number  of  feet  in  the  length  of  a  room  and  the  length  ia 

3  times  the  width,  what  is  the  width?  Aw.  - 

3 

57.  If  there  are  17  books  on  a  shelf  and  7  more  are  added  to  these, 
how  many  are  there?     If  n  books  are  on  the  shelf  and  p  books  are  added, 
how  many  are  there?     If  there  are  q  books  and  t  books  are  taken  away, 
how  many  arc  there  left? 

58.  If  s  is  the  number  of  feet  in  the  length  of  a  line,  how  many  feet 
lire  there  in  twice  this  length?     In  half  the  length?     In  J  the  length? 

59.  A  boy  can  run  a  yards  in  1  second.     How  far  can  he  run  in  10 
seconds  t     In  c  seconds?     In  1  minute?     In  d  minutes? 

60.  If  a.  train  runs  a  miles  in  c  hours,  how  far  does  it  run  in  1  hour?     In 

a  at/ 

(I  hours?  Ans.  -  miles;  -   miles. 

c  c 

61.  A  grocer  receives  c  cents  for  one  pound  of  tea.     How  much  does 
he  receive  for  10  pounds?     For.p  pounds? 

62.  A  boy  can  walk  m  miles  an  hour.     How  many  miles  can  he  walk 
in  c  hours?     In  d  minutes?     How  many  feet  can  he  walk  in  e  minutes? 

md 

Ans.  me;  —  :  Some. 
60 

63.  What  number  does  10(+u  represent  if  t  =  5  and  u=  6?     Ans.  56. 

64.  What  number  does  lOOA  + Wt +u  represent  if  h  -  3,  t  =  7,  and  u  =  9? 
If  h  =  9,  t  =8,  and  w  =  7?  Ans.  379;  987. 

65.  What  number  does  1000a  +  1006  +  10c+d  represent  if  a  =  7,  6  =  8, 
c=2,  and  <J  =  3?  Ans.  7823. 

66.  If  a,  6,  c,  and  d  have  the  same  values  as  in  exercise  65,  what  num- 
ber does  abed  represent?     What  does  7823  mean?  Ans.  336. 

67.  Is  twice  any  whole  number  always  an  even  number?     If  n  is  any 
whole  number,  represent  any  even  number.  Ans.  2n. 

68.  If  n  is  any  whole  number,  does  2n  —  1  represent  an  even  number  or 
an  odd  number?     Does  2n-\-l  represent  an  even  number  or  an  odd 
number? 


CHAPTER  XX 
FORMULAS  AND  TRANSLATIONS 

188.  Subject  matter. — In  the  present  chapter,  drill  will  be 
given  in  evaluating  algebraic  expressions  and  formulas,  in 
translating  verbally  stated  rules  and  principles  into  algebraic 
expressions  and  formulas,  and  in  translating  certain  algebraic 
expressions  and  formulas  into  words. 

In  previous  chapters  many  formulas  have  been  used. 
To  some  extent  the  derivation  of  these  was  made  clear.  But 
one  cannot  thoroughly  understand  the  derivation  of  formulas 
and  the  changes  that  can  be  made  in  them  without  a  consider- 
able knowledge  of  algebra.  It  is  only  through  a  thorough 
understanding  of  the  equation  that  one  gains  the  ability  to 
change  a  formula  to  other  forms  that  are  more  convenient  for 
certain  computations.  The  equation,  in  its  turn,  requires  an 
understanding  of  the  fundamental  operations,  factoring,  and 
fractions  in  algebra  for  its  manipulation.  For  these  reasons 
we  shall  expect  a  more  complete  treatment  of  formulas  later. 

189.  The  slide  rule. — The  slide  rule  is  a  device  consisting 
of  sliding  scales.     It  is   used   to   make  certain  arithmetical 
calculations  in  a  very  simple  manner.     It  gives  the  results  in 
multiplications,  divisions,  squares  and  square  roots,  and  cubes 
and  cube  roots  at  sight,  and  so  saves  time.     However,  the  results 
are  correct  to  only  three  or  four  figures  when  the  common  ten- 
inch  slide  rule  is  used.     For  this  reason,  the  slide  rule  is  to  be 
relied  upon  for  results  only  when  it  is  sufficiently  accurate. 
In   other  cases,  it   may  be  used  to  test  the  results  obtained 
by  ordinary  computation.     In  all  cases,  however,  one  should 
be   able   to  make  the  computations  without  using  the  slide 
rule. 

The  student  is  urged  to  acquaint  himself  with  the  slide 
rule  and  its  uses.  The  computations  in  some  of  the  exercises 
of  this  chapter  may  be  made  by  means  of  this  device.  It  is 
an  invaluable  instrument  in  making  estimates. 

237 


PRACTICAL  MATHEMATICS 

It  is  not  thought  necessary  to  give  a  detailed  description  of 
the  slide  rule  here  nor  to  describe  its  uses,  since  with  each 
instrument  the  buyer  receives  a  booklet  giving  a  full  descrip- 
tion of  the  device  and  its  uses.  At  this  point  in  his  work, 
however,  the  student  cannot  expect  to  fully  understand  the 
uses  of  the  slide  rule  as  it  is  based  in  principle  upon 
logarithms. 

190.  Evaluation  of  algebraic  expressions. — A  numerical 
algebraic  expression  has  a  definite  value  which  may  be  found 
by  performing  the  indicated  operations. 

Thus,  21 -(10+3) +14 -10  =  21 -13 +  14 -10  =  12. 

A  literal  algebraic  expression  has  a  definite  value  depending 
upon  the  values  given  the  letters. 

Thus,  the  expression  abc,  which  means  aX&Xc,  has  a  definite  value  if 
a  =  3,  6=4,  and  c  =  10.  Putting  these  values  in  place  of  the  letters  we 
have  3X4X10  =  120. 

If  any  other  set  of  values  are  assigned  to  a,  b,  and  c,  a  defi- 
nite value  will  be  obtained  for  the  product. 

Example  1.  Find  the  value  of  ?rr2/i  if  r  =  3.1416,  r  =  6,  and 
fc  =  10. 

Substituting  these  values  for  the  letters,  the  expression 
becomes  3.1416X62X  10  =  1130.976,  the  definite  value. 

Definition.  When  an  algebraic  expression  is  the  statement 
of  some  rule  or  principle  it  is  called  a  formula. 

Example  2.  Find  the  value  of  a'+3a26+3a&2+&8,  when 
a  =  2and  6  =  3. 

Substituting  the  values  in  place  of  the  letters, 

a8+3a2&+3a62+68  =  28+3X22X3+3X2X32+38=125, 
the  definite  value. 


Example     2.     Find    the    value  of    Vs(s  —  d)(s— b)(s  —  c), 
if  s=      2         and  0  =  36>  &  =  22,  and  c  =  20. 

o+6+c        36+22+20 
Substituting  in  s=- — ~ — »  s-—  --39. 


Substituting  in  \/s(s  —  a)  (s  —  b)  (s  —  c},  we  have 
•N/39C39-  36)(39-  22)(39-20)  =  \/3779l  =  194.4  - .    Ans. 


FORMULAS  AND  TRANSLATIONS  239 


EXERCISES  66 

1.  Express  without  exponents  and  find  the  value  of  the  following: 
(1)  23-42-                                                 (2)  72-22-5. 

(3)  24-53-72.  Ans.  98,000.     (4)  33  54-72. 

(5)  7-H2-53.                Ans.  105,875.    (6)  34-25-132.  Ans.  438,048. 

If  a  =  2,  6  =  3,  and  c  =  5,  find  the  values  of  the  following: 

2.  3a263.               Ans.  324.                3.  2a62-c2.  Ans.  11. 
4.  3(a2+62).        Ans.  33.                   6.  (a+6)3.  Ans.  125. 
6.  (a+c-6)3.      Ans.  64.                   7.  (c-a)4.  Ans.  81. 

8.  a(a2+c2-62).  Ans.  40. 

9.  a62(c2-62).  Ans.  288. 

10.  (o+6)(c-6).  Ans.  10. 

11.  (c+6+a)(c-6+a).  Ans.  40. 

12.  a2+2a&+62.  Ans.  25. 

13.  (a2+62)2-c2.  Ans.  144. 

14.  c3-(a2+62).  Ans.  112. 

15.  (a2+62)2-(c2-62).  Ans.  153. 

16.  Find  the  value  of  the  following  when  r  =  18: 

(1)  S  =  47rr2.  Ans.  4071.5+. 

(2)  V  =  f^r3.  Ans.  24,429+. 

17.  Find  the  value  of  the  following  when  r  =  7  and  h  =  9  : 

(1)  S=2irrh.  A  ns.  395.84+. 

(2)  V=irrzh.  Ans.  1385.  4  +  . 

If  a  =  1,  b  =  3,  c  =  5,  and  d  =  0,  find  the  numerical  values  of  the  following  : 

18.  a2+262+3c2+4d2.  Ans.  94. 

19.  a4+4a'6+6a262-4a&3+64.  Ans.  40. 


12a3-b2  ,    2c2 

20.  oo         +        i    L->  ri.t          '  AfiS,    O. 

3a2         a+o2          5o3 

Suggestion.  Remember  that  the  lines  in  the  fractions  are  vinculums 
and  bind  the  terms  in  the  numerators  or  denominators  together,  as  well 
as  indicate  division.  The  multiplications,  additions,  and  subtractions 
indicated  in  the  numerators  and  denominators  must  be  performed  first. 
Thus,  after  substituting  the  values, 

12XP-32     2X52     l+32+53  =  3     50     135 
3X1          1+32        5X33        3  +  10     135' 

If  a  =  1,  b  =  2,  c  =  3,  d  =  5,  and  e  =  8,  evaluate  the  following  : 

21.  b2(a2+e2-c2).  Ans.  224. 

22.  (a2+62+c2)(e2-d2-c2).  Ans.  420. 


_ 

23.  e-[v/e  +  l+2+e-v^e-4.  Ans.  15. 
Evaluate  the  following  when  a  =  l,  6=2,  c  =  3,  d  =  4,  and  e~5: 

24.  abc't+bcd2-dea2.  Ans.  94. 

25.  e4+6e262+64-4e36-4e63.  Ans.  81. 


26.  -  ,  ,  0  —  ,,   .  o  —  ,  ,  i  t  -,        '  Ans.  o. 

a3+3a2o+3ao2+o3 

27.  (a+6)(6+c)-(6+c)(c+d)+(c+d)(d+e).  Ans.  43. 


240 


PRACTICAL  MATHEMATICS 


28. 
29. 
30. 


Xru.  72. 
11. 
7. 

1,  6=2,  c-3,  and 
Ans.  6. 


31.  Evaluate  (ac-M)Va*6c~-ffr1c/+c*ad-2,  if  a 
rf-0. 

32.  If  A  stands  for  the  number  of  square  units  in  the  area  of  a  circle, 
and  r  stands  for  the  number  of  linear  units  in  the  radius,  state  in  words 
the  following  formula:  A  =  irrz. 

33.  In  the  above  formula,  can  r  be  any  number  we  please  to  make  it? 
Can  A1     If  we  make  r  =  5  in.,  can  we  then  make  A  anything  we  please? 
Can  T  be  any  number  we  wish  to  make  it  ?     Is  T  a  general  number  ?     Is  A  ? 
Isr? 

34.  Using  A,  a,  and  6,  state  the  following  as  a  formula:  The  area  of 
any  triangle  equals  one-half  of  the  base  times  the  altitude. 

35.  Write  a  formula  for  finding  the  area  of  the  cross  section  of  a 
channel  iron,  using  the  letters  given  in  Fig.  188.      Ans.  A  =  td+b(s+n). 

36.  Using  the  formula  derived  in  the  above  exercise,  find  the  areas  of 
the  cross  sections  of  channel  irons  of  the  following  dimensions: 


FIG.  188. 


FIG.   189. 


No.   d 

t 

b 

n 

8 

Ans.  A 

(1)  3  in. 

0.17  in. 

1.24  in. 

0.38  in. 

0.17in. 

1.19  in. 

(2)  5  in. 

0.19  in. 

1.56  in. 

0.45  in. 

0.19  in. 

1  .  95  in. 

(3)  8  in. 

0.22  in. 

2.  04  in. 

0  56  in. 

0  .  22  in. 

3.35  in. 

(4)  10  in. 

0.24  in. 

2.  36  in. 

0.63  in. 

0.24  in. 

4.45  in. 

(5)  15  in. 

0.40  in. 

3.  00  in. 

0.90  in. 

0.40  in. 

9.90  in. 

(6)  6  in. 

0.41  in. 

3.  15  in. 

0.53  in. 

0.46  in. 

5.58  in. 

(7)  7  in. 

0.45  in. 

3.  00  in. 

0.525  in. 

0.475  in. 

6.  15  in. 

(8)  10  in. 

0.375  in. 

3.  00  in. 

0.469  in. 

0.406  in. 

6.  38  in. 

37.  Write  a  formula  for  finding  the  area  of  the  cross  section  of  an 
I-beam,  using  the  letters  as  given  in  Fig.  189.    Ans.  A  =  dt+'2b(n+s). 

38.  Using  the  same  letters  for  dimensions  as  in  the  last  exercise,  find 
the  areas  of  the  cross  sections  in  the  following  I-beams: 

No.  d  I  b  s  n  Ans.  A 

(1)  5in.       0.21in.       1 .395  in.  0.21  in.    0.44in.       2.86in.» 

(2)  9  in.       029  in.       2. 02  in.     0.29  in.    0.627  in.     6  31  in.* 

(3)  lf>  in.       080  in.       2  80  in.     0  80  in.    1.27  in.     23  59  in.* 

39.  Using  li  for  base,  P  for  percentage,  R  for  rate,  and  A  for  amount, 
write  the  formulas  for  determining  the  different  things  called  for  in 
percentage. 


FORMULAS  AND  TRANSLATIONS  241 

40.  If  p  stands  for  principal,  i  for  interest,  t  for  time  in  years,  r  for  rate 
per  cent,  and  a  for  amount,  translate  the  following  formulas  into  English : 

(1)    i  =  prt.  (2)  a  =  prt+p.  (3)  r.=i+p.  (4)  r  =  -- 

pt 


41.  State  in  words  the  process  of  adding  two  fractions  having  a  common 
denominator.     State  the  same  as  a  formula,  using  a  and  b  for  the  numera- 
tors and  d  for  the  common  denominator.  Ans.    -,+  -.  = — -. — • 

d     d        d 

42.  Do  the  same  as  in  exercise  41,  for  subtracting  fractions  having  a 
common  denominator. 

43.  Translate  the  following  into  English: 

m  a_L_c=a-y(l  C9%  nXm  _n 

(»  b  '  d     be  ^'  dXm~d 

n  +  m     n  n  nXm 

(3) =    •  (4)  -.Xni=-—. — 

d+m     d  d  d 

44.  Write  the  formula  that  states  that  a  times  the  sum  of  b,  c,  and 
d  equals  s.  Ans.  s=a(6+c+d). 

46.  In  exercise  44,  find  s:  (1)  if  a  =  7,  6  =  6,  c  =  8  and  d  =  10;  (2)  if 
a  =  f ,  b  =  |,  c  =  f ,  and  d  =  5V  .  Ans.  168;  l^j. 

46.  Write  in   algebraic   symbols  that  if  c,  the  cost,  in  dollars,  of  a 
harness,  be  increased  by  5,  the  sum  multiplied  by  4  equals  v,  the  cost,  in 
dollars,  of  a  horse.     If  c  is  45,  find  the  value  of  v  in  the  formula. 

Ans.  w  =  4(c+5);  200. 

47.  Using  bi  and  62  for  the  bases  and  h  for  the  altitude,  state  the 
following  as  a  formula:  The  area  of  a  trapezoid  equals  one-half  the  sum 
of  the  two  bases  times  the  altitude.  Ans.  A  =5(^1+^2)^- 

48.  Using  the  formula  of  the  preceding  exercise,  find  the  areas  of  the 
following  trapezoids: 

(1)  61  =22.33  in.,  62  =46.39  in.,  h  =26.43  in.  Ans.  908.13+ in.2 

(2)  6,  =7.203  in.,  62  =  5.826  in.,  h=  3.243  in.  Ans.  21.127-in.2 

49.  Write  the  formula   stating  that  the   area  S  of  the  surface  of  a 
sphere  equals  4  times  IT  times  the  square  of  the  radius  r.     Use  the  for- 
mula and  find  the  area  of  the  surface  of  a  sphere  15  in.  in  radius. 

Ans.  2827.4+  in.2 

60.  Write  a  formula  stating  that  the  volume  F  of  a  sphere  equals  |  of 
IT  times  the  cube  of  the  radius  r.  Use  this  formula  and  find  the  volume  of 
a  sphere  12  in.  in  radius.  Ans.  7328.2+  in.3 

51.  Write  a  formula  stating  that  the  volume  V  of  a  rectangular  solid 
equals  the  length  I  times  the  breadth  6  times  the  height  h.  Use  this 
formula  and  find  the  number  of  cubic  feet  in  a  room  40  ft.  by  30  ft.  by 
12  ft.  Ans.  14,400. 

62.  Find  the  perimeter  of  each  form  in  Fig.  190.  Ans.   (1)  4x; 

(2)  2(a+6);  (3)  2(x+y)+46;  (4)  2(b+c+d+y)+4x;  (5)  2(a+6+d). 
16 


242 


PRACTICAL  MATHEMATICS 


53.  Find  the  area  of  each  form  in  Fig.  190.  An*.  (1)  x1  —  y*; 
(2)  ab-2xy;  (3)  xy-2ab;  (4)  6c-y(2x+d)  -cP;  (5)  ab-cd. 

54.  If  a  man  is  35  years  old,  what  was  his  age  a  years  ago?     What  will 
it  be  6  years  from  now?  Ans.  35— a  years;  35+6  years. 

65.  If  x  years  was  the  age  of  a  man  a  years  ago,  what  is  his  age  now? 
What  will  it  be  in  c  years  from  now?  Ana.  x+a  years;  x+a+c  years. 

66.  Write  in  algebraic  language:  x  diminished  by  a;  y  increased  by  b; 
j  divided  by  n;  the  nth  part  of  x;  one  ruth  of  a. 

67.  The  sum  of  two  numbers  is  a  and  one  of  the  numbers  is  x,  what  is 
the  other  number V     Express  one  with  of  the  first  plus  one  nth  of  the 


second  number. 


x     a-x 

Ans.  a— x;  — |- 

m       n 


vV 

(1) 

*-  —  x  —  »• 

(2) 

..~4 

i 

i  I 

4--*  —  •>> 

If 

] 

*-** 

t-*-* 

V 

i 

" 

i 

-*-x-^»- 

y 

i 
I 

(4) 

FIG.  190. 


68.  If  x  is  the  first  of  two  consecutive  numbers,  what  is  the  second? 
What  will  represent  each  of  three  consecutive  numbers,  if  x  stands  for 
the  middle  one?  Ans.   x  +  1;  x  — 1,  x,  and  x  +  1. 

69.  Represent  three  consecutive  even  numbers  if  x  is  the  middle  one. 
60.  If  the  length  of  a  stick  is  a  feet,  how  many  inches  is  it  in  length? 

How  many  rods?     How  many  miles? 


61.  If  d  is  the  number  of  dollars  an  article  costs,  what  is  the  number 
of  cents?  Ans.  lOOd. 

62.  If  n  room  is  a  yards,  6  feet,  and  c  inches  long,  how  many  inches  in 
length  is  it?  Ans.  36a  +  126+c. 

63.  Given  the  product  p  and  the  multiplier  m,  find  the  multiplicand  n. 

Ans.  n=— • 


FORMULAS  AND  TRANSLATIONS  243 

64.  Given  the  dividend  d  and  the  quotient  q,  find  the  divisor  6. 

d 

Ans.  b=-- 
Q 

66.  Given  the  divisor  d,  the  quotient  q,  and  the  remainder  r,  find  the 
dividend  D.  Ans.  D=dq+r. 

66.  The  difference  between  two  numbers  is  n  and  the  smaller  one  is  a, 
what  is  the  larger? 

67.  A  number  exceeds  a  by  c,  what  is  the  number?  Ans.  a+c. 

68.  Write  in  symbols  that  a  exceeds  b  as  much  as  c  is  less  than  d. 

Ans.  a  —  b  =d  —  c. 

69.  Express  in  symbols  that  one-half  of  m  equals  the  nth  part  of  the 

f      T  i        a-\-b-\-c 

sum  of  a,  b,  and  c.  Ans.  \m=  -- 

70.  If  a  cents  is  the  price  per  quart  for  beans,  what  is  the  price  per 
bushel?  Ans.  32a  cents. 

71.  The  provisions  that  will  keep  a  family  of  9  persons  30  days  will 
keep  a  family  of  5  persons  how  many  days?     The  provisions  that  will 
keep  a  family  of  a  persons  I  days  will  keep  6  persons  how  many  days? 

_.     al 
Ans.  54;   T- 

72.  If  it  takes  m  men  d  days  to  dig  a  ditch,  how  many  days  will  it  take 

md 

c  men  to  dig  it:  Ans.  -- 

C 

73.  How  many  pounds  of  sugar  at  c  cents  a  pound  will  d  dozen  eggs  at 
e  cents  a  dozen  buy?  Ans.  —  • 

74.  A  cubical  tank  e  ft.  on  an  edge  will  hold  how  many  barrels,  if  one 

e3 
barrel  equals  4.211  cu.  ft.?  Ans.  .         • 

4.^11 

75.  A  box  a  ft.  long,  b  ft.  wide,  and  c  ft.  deep  will  hold  how  many 
bushels,  if  1  bushel  equals  2150.42  cu.  in.?  Ans. 


76.  A  man  earned  $a  per  day  and  his  son  $6.  How  many  dollars  did 
they  both  earn  in  a  month,  if  the  man  worked  26  days  and  the  son  21 
days?  Ans.  26o+216. 


CHAPTER  XXI 
POSITIVE  AND  NEGATIVE  NUMBERS 

191.  Meaning  of  negative  numbers. — The  degrees  of  tem- 
perature, indicated  by  the  thermometer  scale,  are  counted  in 
two  opposite  directions  from  the  zero  point.     We  usually 
speak  of  a  temperature  as  so  many  degrees  above  or  below 
zero.     In  arithmetic  we  speak  of  temperature  in  this  manner, 
but  in  algebra  we  seek  some  abbreviated  form  for  stating  the 
same  thing.     We  might  agree  to  use  any  convenient  signs 
whatever.     The  signs  +  and  —  have  been  generally  adopted. 
The  +  sign,  placed  before  the  number  of  degrees,  indicates  a, 
temperature  above  zero;  and  the  —  sign  indicates  a  tempera- 
ture below  zero.     This  use  of  these  signs  is  different  from  the 
ordinary  use  in  which  they  indicate  addition  and  subtraction. 
Here  they  indicate  the  sense  or  direction  in  which  the  tempera- 
ture is  measured  or  counted. 

Thus,  +25°  means  25°  above  zero,  and  —25°  means  25°  below  zero. 

The  number  preceded  by  the  +  sign  is  called  a  positive 
number,  and  the  one  preceded  by  the  -  sign,  a  negative 
number.  Two  numbers  so  related,  one  positive  and  one 
negative,  may  be  called  relative  numbers.  The  following 
are  further  examples  of  relative  numbers.  They  will  help 
to  fix  the  negative  number-idea  in  mind. 

Time  is  commonly  measured  forward  and  backward  from  a 
certain  date.  This  might  be  shown  by  using  the  +  and  —  signs. 

Thus,  1918  A.  D.  might  be  written  +1918,  while  325  B.  C.  might  1>c 
written  —325. 

A  force  acting  in  one  direction  and  another  in  an  opposite 
direction  are  designated  as  a  +  and  a  —  force  respectively. 
Money  gained  and  resources  are  + ,  and  money  lost  and  lia- 
bilities are  — . 

192.  Need  of  negative  number. — The  necessity  for  extend- 
ing the  number  system  so  as  to  include  negative   numbers 
may  be  .seen  from  the  following  subtractions,  where  the  min- 

244 


POSITIVE  AND  NEGATIVE  NUMBERS  245 

uend  remains  the  same,  but  the  subtrahend  increases  by  steps 
of  1  as  we  pass  from  left  to  right.  This  causes  the  difference 
to  diminish  by  steps  of  1  from  left  to  right.  When  the  differ- 
ence becomes  less  than  zero  we  indicate  it  by  the  sign  — 
placed  before  the  number. 

6666666         6 
_3     _4     _5     J>  _JT  _8  __9     JO 
3       2       1       0-1-2-3-4  etc. 

193.  Representation  of  negative  and  positive  numbers. — 

For  convenience,  the  positive  and  negative  numbers  may 
be  represented  on  a  horizontal  line;  the  +,  or  positive,  numbers 
to  the  right  of  a  certain  point,  called  zero,  and  the  — ,  or  nega- 
tive, numbers  to  the  left  of  the  zero  point.  This  method  of 
representing  them  is  found  very  convenient  in  explaining  addi- 
tion and  subtraction. 

It  should  be  carefully  noted  that  toward  the  right  is  the 


— t -6-5-4-3-2-1    0  +  1+2  +  3+4  +  5  +  6 ^+ 

i     i     i     i     i     i     I     i     i     i     i     i     i 

FIG.   191. 

positive  direction  and  toward  the  left  is  the  negative  direction, 
no  matter  from  what  point  we  start. 

The  idea  of  negative  number  is  opposite  to  that  of  positive 
number.  For  example,  if  a  man  walks  five  miles  east,  or  in 
the  positive  direction,  and  then  five  miles  west,  or  in  the  nega- 
tive direction,  he  is  at  the  starting-point.  The  negative 
distance  has  destroyed  the  opposite  or  positive. 

194.  Definitions. — Positive  and  negative  numbers,  to- 
gether with  zero,  form  the  system  called  algebraic  numbers. 

The  absolute  or  numerical  value  of  a  number  is  the  value 
which  it  has  without  reference  to  its  sign. 

Thus,  +5  and  —5  have  the  same  absolute  value  5. 

The  signs  +  and  --  when  used  to  show  direction  or  sense 
are  called  signs  of  quality  to  distinguish  them  from  the  signs 
of  operation  used  to  indicate  an  addition  or  a  subtraction. 

The  sign  +,  used  as  a  sign  of  quality,  is  usually  omitted; 
but  the  sign  — ,  when  a  sign  of  quality,  is  expressed. 

To  show  that  the  sign  is  one  of  quality  it  is  sometimes 


246  PRACTICAL  MATHEMATICS 

written  with  the  number  and  enclosed  in  parentheses,  as  (—3), 
(+4).  (-5)  +  (+2)  indicates  that  a  -5  is  to  be  added  to 
a +2. 

196.  Remarks  on  numbers. — What  is  the  negative  number 
and  why  do  we  need  to  trouble  ourselves  about  it?  The 
illustrations  given  above  should  help  one  to  get  the  idea  of 
negative  number,  and  to  see  how  it  is  forced  upon  us  when 
we  try  to  subtract  a  larger  number  from  a  smaller  one. 

In  mathematics  when  a  new  number-idea  appears,  the  first 
thing  to  do  is  to  represent  it  by  a  symbol;  and,  second,  find 
a  way  of  operating  with  it.  That  is,  we  must  determine  how 
to  add,  subtract,  multiply,  and  divide  such  numbers. 

One  of  the  first  things  we  did  in  arithmetic  was  to  determine 
methods  of  operating  with  positive  whole  numbers;  then  a 
little  later,  we  did  the  same  thing  for  the  fractional  numbers. 
In  fact,  much  of  the  time  spent  in  studying  mathematics  is 
spent  in  finding  how  to  add,  subtract,  multiply,  and  divide 
numbers  of  different  kinds,  whole,  positive,  negative,  fractional 
and  combinations  of  these. 

It  now  remains  to  devise  methods  and  rules  for  operating 
with  algebraic  numbers.  This  must  be  done  in  such  a  way  that 
no  old  rules  or  principles  shall  be  violated.  The  student  is 
asked  to  consider  carefully  each  step  taken  and  to  make  every 
part  seem  reasonable. 

EXERCISES  56 

1.  A  place  is  at  10°  25'  east  longitude.     What  sign  may  be  used  instead 
of  the  word  east  to  show  this? 

2.  New  York  is  74°  west  longitude.     Designate  this  by  means  of  a 
sign. 

3.  Chicago  is  41°  45'  north  latitude,  and  Melbourne  37°  south  latitude. 
Designate  these  algebraically,  that  is,  with  signs. 

4.  A  man  overdraws  his  bank  account.     How  may  the  banker  indicate 
this  on  his  books? 

5.  If  the  weight  of  a  stone,  weighing  100  lb.,  is  written  4-100  lb.,  how 
may  the  upward  lift  of  500  lb.  of  a  balloon  be  written? 

6.  What  is  meant  by  saying  that  a  man  is  worth    +$3000?     By 
-$2000? 

7.  If  a  man  has  a  -$200,  a  +$300,  and  a  -$100,  what  is  he  worth? 

8.  A  man  was  born  in  the  year  —35  and  died  in  the  year  +48.     How 
old  waa  he  when  he  died?  Ans.  83  years. 


POSITIVE  AND  NEGATIVE  NUMBERS  247 

9.  If  a  man  has  assets  of  $3000,  and  we  represent  it  by  a  line  3  in.  long 
drawn  to  the  right  of  0,  how  could  we  represent  a  liability  of  $1500? 

10.  A  man  walks  10  miles  east  and  then  6  miles  west.     Write  this  in 
symbols. 

11.  If  a  body  is  heated  37°  and  then  cooled  down  42°,  indicate  the 
facts  in  symbols. 

12.  If  7  men  came  into  a  room  and  8  men  left  the  room,  we  could  rep- 
resent the  fact  by  saying  that  +7  and  —8  men  came  into  the  room.     A 
man  received  $25  and  spent  $21.     Write  this  as  money  received. 

13.  Remembering  the  positive  and  negative  directions,  draw  a  line; 
locate  a  zero  point;  and,  with  a  unit  of  1  in.,  locate  positive  and  negative 
numbers  to  the  right  and  left  of  zero.     Start  at  +5  and  go  +3,   -  7, 
-4,  +6,  -10,  +2,  +1,  and  +4.     Where  do  you  stop? 

Ans.  At  the  zero  point 


CHAPTER  XXII 
ADDITION  AND  SUBTRACTION 

196.  Definitions. — The   aggregate   value   of   two   or   more 
algebraic  numbers  is  called  their  algebraic  sum.     The  process 
of  finding  this  sum  is  called  addition. 

197.  Addition  of  algebraic  numbers. — If  we  wish  to  add 
3  to  4,  we  start  with  4  and  count  3  more,  arriving  at  7  which 
is  the  sum.     If  we  consider  the  system  of  algebraic  numbers 
arranged  on  a  horizontal  line,  Fig.  192,  we  add  a  positive  num- 
ber by  starting  with  the  number  we  wish  to  add  to,  and  count- 

-10-9-8-7-6-5-4-3-2-1    0+1+2 +3*4+5 +6+7 +8+9+10 

i     i     i     i     i     i     i     i     i     i     I     i     i     i     i     i     i     i     i     i     i 

FIG.  192. 

ing  toward  the  right  as  many  units  as  there  are  in  the  number 
added. 

Thus,  in  the  above,  we  start  at  4  and  count  toward  the  right  to  7. 

To  add  +5  to  —3,  we  start  at  —3  and  proceed  5  units  toward  the 
right,  arriving  at  +2. 

To  add  +3  to  —7,  we  start  at  —7  and  proceed  3  units  toward  the 
right,  arriving  at  —4. 

Since  adding  —3  to  +7  is  the  same  as  adding  +7  to  —3, 
the  result  of  adding  —3  to  +7  is  +4.  In  order  to  start 
with  +7  and  arrive  at  +4,  we  must  move  in  the  negative 
direction,  or  toward  the  left.  Therefore,  we  conclude  that 
to  add  a  negative  number, we  go  toward  the  left. 

Thus,  to  add  —4  to  +9,  we  start  at  +9  and  go  4  units  toward  the 
left,  arriving  at  +5.  To  add  —7  to  +2,  we  start  at  +2  and  go  7  units 
toward  the  left,  arriving  at  —5.  To  add  —4  to  —5,  we  start  with  —5 
and  proceed  4  units  toward  the  left,  arriving  at  —9. 

The  above  results  arc  given  here: 

+  4  -:<  -7  +7  +9  +  2  -5 
+  :*  +1*  +  3  -3  -4  -7  -4 
+  7  +2  -4  +4  +5  -5  -9 

248 


ADDITION  AND  SUBTRACTION  249 

198.  Principles. — A    careful    consideration    of    the    above 
will  disclose  the  following  principles: 

(1)  The  algebraic  sum  of  two  numbers  with  like  signs  is  the 
sum  of  their  absolute  values,  with  the  common  sign  prefixed. 

(2)  The  algebraic  sum  of  two  numbers  with  unlike  signs  is 
the  difference  between  their  absolute  values,  with  the  sign  of  the 
one  greater  in  absolute  value  prefixed. 

In  adding  three  or  more  algebraic  numbers,  differing  in 
signs,  find  the  sum  of  the  positive  numbers,  and  then  the  sum 
of  the  negative  numbers  by  principle  (1),  and  then  add  these 
sums  by  principle  (2). 

Thus,  in  finding  the  sum  of 

+2,  +10,  -6,  -3,  -7,  +9,  we  take 
+2  +  10+9  = +21  and 
(-6)  +  (-3)  +  (-7)  =  -16,  then 
+21 +  (-16)  =  +5,  the  sum. 

199.  Subtraction  of  algebraic  numbers. — Subtraction  is  the 

inverse  of  addition.  If  we  are  given  one  of  two  numbers  and 
their  sum,  subtraction  is  the  process  of  finding  the  other  number. 

In  arithmetic  it  is  assumed  that  the  minuend  is  always 
greater  than  the  subtrahend.  In  the  subtraction  of  algebraic 
numbers  we  not  only  may  have  the  subtrahend  larger  than 
the  minuend  when  the  numbers  are  positive,  but  either  or 
both  subtrahend  and  minuend  may  be  negative  numbers. 

Since  subtraction  is  the  inverse  of  addition,  if  we  consider 
the  system  of  algebraic  numbers  arranged  along  the  horizontal 
line  as  in  Art.  197,  we  have  the  following  principles: 

(1)  Subtracting  a  positive  number  is  equivalent  to  adding 
a  numerically  equal  negative  number. 

(2)  Subtracting  a  negative  number  is  equivalent  to  adding  a 
numerically  equal  positive  number. 

These  may  be  combined  in  the  following: 

RULE.  Subtraction  of  algebraic  numbers  is  performed  by 
considering  the  sign  of  the  subtrahend  changed  and  proceeding 
as  in  addition  of  algebraic  numbers. 

Applying  the  rule,  we  find  the  following  algebraic  differences : 

+  7  +4  -6  -3  -8  +  7 
+  3  +6  -2  -7  _+3_  -2 
+4  -2  -4  +4  -11  +9 


250  PRACTICAL  MATHEMATICS 

It  should  be  carefully  noted  that  4  —  3  may  be  considered 
as  a  +4  minus  a  +3,  or  as  a  +4  plus  a  —3.  It  is  this  choice 
that  causes  more  or  less  trouble  to  the  beginner. 

If  the  number  with  its  sign  of  quality  is  inclosed  in  paren- 
theses, we  have,  for  example,  (+4)  +  (  —  6)  —  (+7)  —  (  —  11)  + 
(+7).  This  may  also  be  written  (+4)  -(+6)  -(4-7)  + 
(  +  ll)  +  (+7),  which  is  the  same  as  4  —  6  —  7+11+7  where  the 
signs  indicate  operations  and  all  the  numbers  are  positive. 

EXERCISES  57 

Find  the  sum  in  exercises  1  to  7. 

1.  7,  -10,  -13,  16,  25,  -3.  Ans.  22. 

2.  27,  46,  -100,  -16,  17.  Ans.   -26. 

3.  3,  16,   -21,   -1,  2,  1.  Ans.  0. 

4.  !,  -f,  I,  -i"«.  An*.  «. 
6.   -3},  -7A,  3|,  -A.  Ans.   -8|J. 

6.  3.25,  -7.16,  -10.3,  14.1.  Ans.   -0.11. 

7.  14.17,  -16.19,  -26.3.  Ans.   -28.32. 

8.  From  17.6  take  -14.3.  Ans.  31.9. 

9.  From  -111  take  -12.  Ans.    -99. 

10.  From  -46  take  75.  Ans.   -121. 

11.  Is  the  absolute  value  of  an  algebraic  number  ever  increased  by 
subtraction?     Illustrate. 

12.  Is  the  absolute  value  of  an  algebraic  number  ever  decreased  by 
addition?     Illustrate. 

13.  How  many  degrees  of  latitude  between  places  at  +37°  45'  17" 
and  at  -16°  14'  53".  Ans.  54°  0'  10". 

14.  If  a  steamer  is  moving  through  still  water  at  the  rate  of  20  miles 
per  hour,  and  a  man  walks  forward  on  the  deck  at  the  rate  of  4  miles 
per  hour,  express  the  rate  the  man  is  moving  with  reference  to  the  water. 
Suppose  the  man  walks  toward  the  stern  of  the  boat  at  the  same  rate, 
how  may  the  rate  of  the  boat,  and  the  rate  of  the  man  with  reference 
to  the  boat,  be  expressed? 

200.  Addition  and  subtraction  of  literal  algebraic  expres- 
sions. —  We  add  5  bushels,  8  bushels,  10  bushels  and  get  23 
bushels.  So  we  have  5  bu.+8  bu.+lO  bu.  =  23  bu. 

Similarly,  6d+4d+7d=17d, 

4xy-\-7xy+8xy  =  19xy,  and 


In  subtraction,  we  have  17o—  5a  =  12a  and 

46z  V  -  6z  V  -  40z  V- 
We  know  that  in  arithmetic  we  cannot  add  or  subtract 


ADDITION  AND  SUBTRACTION  251 

unlike  things;  neither  can  we  do  so  here.     If  we  wish  to  add 
3a  to  26,  we  indicate  the  addition,  thus,  3a+26. 

From  these  considerations,  we  have  the  following  principle: 
Monomials  which  are  alike,  or  similar,  can  be  added  or  sub- 
tracted by  adding  or  subtracting  the  coefficients.     If  the  mono- 
mials are  unlike,  the  operations  can  only  be  indicated. 

Examples  of  addition. 

(1)  (2)  (3) 

+  3abc  —  Wxy3  I7ab 

-  Qabc  +  3xy3  —  3xy 

+  10a6c  -  4xy3  -4c2 

-16a6c  -  7xy3  +3a2 
-  3a6c 


-I2abc                +  4xy3  17a6-3:n/-4c2+3a2 
Examples  of  subtraction. 

(1)                         (2)  (3) 

-21x2y  Uab 

3xzy  —  6c 

-24x2y  14a6  +  6c 

201.  Polynomials.  —  The  addition  and  subtraction  of  poly- 

nomials is  similar  to  that  of  monomials.     Write  them  so  that 

like  terms  are  in  the  same  column,  and  combine  the  terms  in 
each  column  as  with  monomials. 

Example  of  addition.  Exam.ple  of  subtraction. 

-   32  17a:y2-14c2+4a 

+  7z  IQx2-  5c2-8a 


9z  7rn/2-9c2+12a 


ax2+  4?/2+  2z 

202.  Test  or  proof  of  results.  —  It  is  very  important  that  one 
should  be  able  to  test  the  results.  The  problems  in  addition 
or  subtraction  of  literal  algebraic  expressions  may  be  tested  by 
substituting  some  definite  values  for  the  general  numbers. 

Thus,  if  a  =  l,  &  =  1,  and  x  =  \  in  the  following  example,  the  test  is  as 
given. 

Example.  Test. 

-  7ab+  422-3&Z  -  7+  4-3=-  6 

-  Sa&-10z2-4&z  -  8-10-4=  -22 

lx2+6bx  ~  9  +  11+6=       8 


-*>4ob+  5x2-   bx  -24+  5-1  =  -20 


252 


PRACTICAL  MATHEMATICS 


The  test  depends  upon  the  fact  that  the  letters  used  may 
have  any  values  whatever.  We  could  just  as  well  take  o  =  l, 
6  =  2,  x  =  3.  Of  course,  we  usually  choose  values  that  make  the 
computations  as  simple  as  possible. 


Ans.  24a. 

Ans.  3ab. 

Ana.   -8x*. 


EXERCISES  68 

Write  the  sums  in  exercises  from  1  to  11. 

1.  3a,  5a,  7  a,  9a. 

2.  6a6,  -4a6,  3a6,  -2ab. 

3.  9x»,  10x»,   -14x»,  -13x*. 

4.  14xV,  -llxV,  -16xV- 
6.  2w+n,  6/n  —  4n,  7m—  2n. 

6.  p+2q-r,  2p-3g. 

7.  a+46-6c,   -26+5c. 

8.  —  2x+3y  —  4z,  x+y—z. 

9.  2x-|-3a-r-w,  2y-3a-m. 

10.  a-(-3c-17,  5a-6c  +  14,  2a-5c-2. 

11.  2a'-4cd,  -8a2-7cd,  -25az  +  16cd. 

12.  Subtract  exercises  6  to  9  above. 

13.  From  3ox-4cd  take  10ox-2cd. 

14.  From  17o6-2c  take  3<zb-4d. 

16.  From  m2  —2mn+n2  take  m*+2mn+nI. 

16.  From  16xV«+4a  take  16a-14cd. 

17.  Add  and  test  by  putting  x  =  l,  y=2,  z  =  3:  14x-16j/-r-3z,  7x+2y- 
7z,  -16z-2y+4z. 

18.  Add   and  test  by  putting  x  =  l   and   j/=2:   3x+2j/—  4xy,   7x  — 


Ans.  15m—  5n. 

-An*.  3p-g-r. 

/Irw.  a+26-c. 

Am.   —  x+4y  —  5z. 

Ans.  2x+2y. 

Ans.  8a-8c-5. 

XTW.    -31a«+5cd. 

Ans.   -7ox-2cd. 

Ans.  14o6-2c+4<f. 

Ans.    —  4mn. 

.  16xVz- 


19.  Subtract  and  test  by  putting  x  =  2,  y  =  1,  and  z  =  3:  3xy—  4xz  —  3x* 
from  —  6xi/+5jz  —  2yl. 

20.  From  17xV-16xzs  +  14clJ  take  -16xV-M3cld-14xz3. 

^  ns.  33x  V  -  2xz»  -  29 

21.  Add  3«6-2ar+4d/,  7ac-6J/,  4a6+9ac-3<//. 

Ann. 

22.  Add  7x'-9i/l-llxi/,  10x*-4xj/-lly*+7x». 

Ans.  24x»-20z/*-15xy. 

23.  Add  llxV-16xy  +  14ax*-17x,  16xV  +  14xy,  17xV-43x+27ax». 

Ans.  44xV-2xt/+41ax»-60x. 

24.  Add  x»+2xj/+y»,  x»-2xj/+j/»r   -2x*+2j/»,  7x»-12i/1. 

Ans.  7xt-Syt. 

26.  Add  34ax-756j/+60rz,  16ax+256i/-10cz,  -41ax+4l6y-20rz. 

Ans.  9ax-96y-f30cz. 
26.  Add  142x4+31xJy+9xV  +  10xys-157/4,   -130x4-30x»y-f  2xV~ 


27.  Add  and  test  by  letting  x  =  l,  y  =  2,  and  z  =  3:  17x-9y,  3z  +  14x, 
-3x,  x-\lz,  x-3y+4z. 


ADDITION  AND  SUBTRACTION  253 

28.  Add  and  test  by  letting  x  =  1,  and  y  =  2:  3x*+2xy+4y2,  4z2-3xy- 
2y2,  3x2+xy. 

29.  Add  and  test  by  letting  m  =  l,  n  =  l,  p==2,  and  g=4:  16m+3n  —  p, 
p+4q,  —  q+7m—3n,  n—q,  3n+2p. 

30.  Add  Sax—  4ab  —  xy+z  —  3xy+  Gab  -\-7xy  —  3z  —  4ax  —  Qxy+ax. 

Ans.  2ab-3xy-2z. 

31.  Add    17ax2-9axy+6z-24+G+2axy-2z+ax2-13-Uax2+6axy 
—  4z.  Ans.  4ax2—axy—3l. 

32.  Subtract  4w2-  6n3  +73x  from  -m2-8n3+83.c. 

Ans.    —  5m2—  2n3  +  10a;. 

33.  Subtract  lQx2y-4xy  +  lSy*  from  3xy-18y2. 

Ans.    —  Wx2y+7xy-34y2. 

34.  From  ax-  +3a?/2  —  4z2  subtract  2ax2+3ay2  —  4z2.          Ans.    —ax2. 
36.  From  5o6c+3bcd+7cde  take  4abc-10bcd-8cde. 

Ans.  abc  -\-13bcd  -\-\5cde. 

36.  Take  2x3  —  y2  from  the  sum  of  x3  —  2xy-{-3y2  and  xy-\-4y2. 

Ans.    —  x3  —  xy  -\-8y2. 

Given  ^1  =a3+3a26+3a&2+63,  B  =  -3a2b+3ab2-3b3,  and  C=a*-bs, 
find  results  in  exercises  37  to  44.  Test  results  by  substituting  values 
for  a  and  b. 

37.  A-B.  38.  B-C.  39.  C-A. 
40.  5  -A.                     41.  C-B.  42.  A-C. 
43.  A+B+C.              44.  A+5-C. 

46.  From  the  sum  of  0.4x2  -7.5a  +5m2  and  1  -0.125a+3m3  take  1.5a  — 
7.2x2  -3.25m3.  Ans.  7.6x2-9.125a+5m2+6.25m3  +  l. 

46.  Find  the  sum  of  io62-fcd3  +  fx3,  f  ab2  -  led3  - 
fx3,  and  fa62  +  fcd3+2x3.  Ans.  2 


203.  Terms  with  unlike  coefficients.  —  It  often  happens 
that  we  wish  to  add  or  subtract  terms  where  the  coefficients 
that  are  to  be  united  are  not  all  numerical.  For  example, 
add  d*x,  e*x,  and  ex  by  uniting  the  coefficients  of  x.  Here  the 
coefficients  of  x  are  d2,  e2,  and  c.  Since  these  are  unlike  terms 
the  addition  can  only  be  indicated;  thus,  d2+e2+c.  We  may 
write  the  sum  then  of  d*x,  e2x,  and  ex  as  (dz-\-ez-\-c}x.  Simi- 
larly the  sum  of  6x,  5x,  and  2x  may  be  written  (6+5+2)x; 
but  here  the  coefficients  can  actually  be  united  and  expressed 
as  one  symbol,  thus  13:r. 

EXERCISES  69 

Add  the  following  by  uniting  the  coefficients  of  the  letter  that  is  com- 
mon in  the  terms. 

1.  5ay,  -6dy,  ley,  and  lly.  Ans.  (5a-&d+4c  +  17)y. 

2.  3x*y,  -llxy,  6x3y,  and  7y.  Ans.   (3x2-14x+6x3+7)y. 

3.  bxy,  -bn,  -146,  and  3bx2y.  Ans.  (xy-n-U+Sx^b. 


254  PRACTICAL  MATHEMATICS 

4.  (a+b)d,cd,  -dfg,  and  2d.  Ant.  (a-H>+c-/0+2)<f. 

6.  (3x-y)n4,  *»n4,  and  (2z+3j/)n«.  Ans.  (z*+5z+2y)n4. 

6.  (a+b-c)x,  (a-b+c)x,  and  (2a-36+4c)z. 

4n«.  (4a-36+4c)z. 

7.  (ac-cd+V)y,  (2ac-3cd-b)y,  and  (4ac-Qcd-4b)y. 

Ans.  (7ac-\Qcd—4b)y. 

Subtract  the  second  expression  from  the  first  in  the  following,  uniting 
the  coefficients  of  the  common  letters. 

8.  ay  and  by.  Ana.   (a  —  b)y. 

9.  3o6and76c.  Ans.  (3a-7c)6. 

10.  4a»z  and  962x.  Ans.  (4a»-96J)z. 

11.  (a+6)zand  (a-2b)z.  Ans.  3bz. 

12.  (2a-36+c)deand  (a+b-2c)de.  Ans.  (a-4b+3c)de. 

13.  (4z+2y  -3z)ab  and  (3i-4y  +2z)ab.  Ans.   (x+6y-5z)ab. 

14.  (ab+cd-ef)x&nd  (ab-2cd-3ef)x.  Ans.  (3cd+2ef)x. 

15.  Find  the  perimeter  of  an  irregular  hexagon  having  sides  of  the 
following  lengths:  2a+b,  3a-26,  4a+b,  2a-36,  2a+26,  and  la-  56. 

Ans.  20a-66. 

204.  Signs  of  grouping.  —  When  a  sign  of  grouping  is  pre- 
ceded by  a  +  or  —  sign,  it  indicates  that  the  expression  en- 
closed by  the  sign  of  grouping  is  to  be  added  to  or  subtracted 
from  what  precedes. 

When  a  plus  sign  precedes  a  sign  of  grouping,  we  may  remove 
the  sign  of  grouping  without  making  any  change  in  signs. 

Thus,   a+(6-c)  =a+b-c. 

When  preceded  by  a  minus  sign,  the  sign  of  grouping  may 
be  removed  if  the  signs  within  it  are  changed. 
Thus,  a-(b-c+d)=a-b+c-d. 

The  reason  for  this  change  is  the  same  as  for  the  changing 
of  the  signs  in  the  subtrahend  when  subtracting. 

When  there  are  several  signs  of  grouping,  one  within  another, 
they  may  be  removed  by  first  removing  the  innermost  one, 
and  then  the  next  outer  one,  continuing  till  all  are  removed. 

Example  1.     Simplify  4x2-5y2+x-[Qx2—3x—(y*—x')]. 

Beginning  with  the  inner  sign  of  grouping, 

4z2  -  by  2  +  x  -  [6x2  -  3x  -  (y*  -  z)  ] 


=  4z2  -  5?/2  +  z  -  6z2+  3:r  +  y2  -  z 

-2z2-4j/2+3z.    Ans. 
Example  2.     Simplify  8  -  {  7  -  [4  +  (2  -  z)J  }  . 


ADDITION  AND  SUBTRACTION  255 


Solution.     8-  {7-[4  +  (2-s)]}  =8-  {7-[4  +  2-x]\ 

=  8-{7-4-2+z} 
=  8-7+4+2-0: 
=  7  —  x.     Ans. 

The  terms  may  be  united  as  soon  as  like  terms  appear 
within  a  sign  of  grouping. 

Thus,  in  the  first  step  in  example  2,  the  4+2  within  the  signs  [  ]  may 
be  united;  it  then  would  be: 

8-{7-[6-z]} 
=  8-{7-6+z} 


=8-1-3 

=  7—  x.     Ans. 

EXERCISES  60 

Simplify  by  removing  the  signs  of  grouping  and  uniting  the  like  terms 
in  the  following: 

1.  4a+76-(3a+26).  Ans.  a+56. 

2.  5-3z  +  (-18+2z).  Ans.    -13-rr. 

3.  llx  +  l-(-x+3).  Ans.   12x-2. 

4.  a-3a2+7-(2a2+5-3o).  Ans.  4o-5a2+2. 
6.  x3  —  z3  —  (x3+?/3  —  z3).  Ans.    —  7/3. 

6.  a—  c-\-d  —  (a+c—  d)  —  (d  —  c  —  a).  Ans.  a  —  c+d. 

7.  3z-(t/-2a;)+(z+?/-52;).  Ans.  2. 

8.  z  —  [y  —  (z—x)}.  Ans.  2z  —  x  —  y. 

9.  74  -26  --(15  -8).  Ans.  41. 

10.  63  -[23  -(14  -8)].  Ans.  46. 

11.  4o+[2o-(o+6)+6].  Ans.  5a. 

12.  3x-[2y+5z-(3x+y)].  Ans.  Qx-y-5z. 

13.  a—  [a—  {a  —  (2a—  a)}].  Ans.  0. 

14.  Let  a  =7,  a;  =  10,  y  —  —  5,  and  z  =  —  2,  and  verify  results  in  exercises 
3,  4,  8,  and  12. 

16.  a+b-[a-b  +  {a+b-(a-b)}}.  Ans.  0. 

16.  a-26-[3o-(6-c;-5c].  Ans.  4c-6-2a. 

17.  2o-{6-[3a  +  (26-o)]}.  Ans.  4a+6. 

18.  5a;-(4x-[-3x-{2x-a;-l}]).  Ans.   l-3x. 

19.  24x23x-2xQx*-x2-4:x+x2.  Ans. 


205.  Insertion  of  signs  of  grouping.  —  For  the  same  reasons 
as  given  in  the  preceding  articles,  any  terms  of  a  polynomial 
may  be  enclosed  in  a  sign  of  grouping  preceded  by  a  plus 
sign  without  change  of  signs.  They  may  be  enclosed  in  a  sign 
of  grouping  preceded  by  a  minus  sign,  provided  the  sign  of  each 
term  within  is  changed  from  —  to  +  ,  or  from  +  to  —  . 


j:,t,  PRACTICAL  MATHEMATICS 

Example.     Enclose  the  last  three  terms  in  the  following 
expression   within   parentheses:  (1)   preceded  by  a   +   sign, 
and  (2)  preceded  by  a  —  sign. 
ax+by+cd— e. 

(1)  ax+by+cd— e  =  ax+(by+cd  —  e). 

(2)  ax -\-lnj -\-cd- e  =  ax  —  (  —  by  —  cd+e). 

EXERCISES  61 

Insert  parentheses  around  all  the  terms  that  follow  the  first  —  sign  in 
each  of  the  following: 

I.  a-b+c.  2.  2a+36-c-2d. 

3.  4z-zl+2/2-a:y.  4.  \7xt—x3y—xy+yz. 

5.  G+3x-4x*+5x3-6x4.  6.   -a-b+c-d+e. 
Write  the  following  with  the  last  three  terms  of  each  enclosed  in 

parentheses:  (1)  preceded  by  a  —  sign,  and  (2)  preceded  by  a  +  sign. 

7.  3x-2y+4-7z.  8.  9x*+4y*-7xy-4yz. 

9.  a+6+c-3/+4c*.  10.  8z+4zy-7z+4. 

II.  4d*+3cJ-4z+3i/.  12.  2c-4dx*+3y-7. 

Collect  all  the  coefficients  of  x  in  the  following  within  parentheses 
preceded  by  a  —  sign. 

13.  ax  —  bx+cx.  Ans.    —  (—  a-\-b—  c)x. 

14.  2cx-4dx+6ex-2x.  ll     Ana.   -(-2c+4d-Ge+2)x. 

15.  x+3ox-4cx+dx.  Ans.    -(-1  -3a+4c-d)x. 

16.  —  x+Qcx  —  4a2x—  ac2x.  Ans.    —  (1  —  6c+4aI+ac*)x. 

In  the  following,  group  within  parentheses  the  terms  that  have  the 
same  letter  to  the  same  power:  (1)  preceded  by  a  +  sign,  (2)  preceded 
by  a  —  sign. 

17.  3-ofe-zy+oc— xz+od.  Ans.  (1)  3 +  ( -a 
(-xy-xz),  (2)  3-(ob-oc-ad)-(xy+xz). 

18.  az—xy+ad  —  ac+px. 

19.  ab1—  act+adt—  nmt+npt  —  nq*. 

20.  acx1  —  bcy+4ncd— 6+4z  —  3j/J. 

21.  7 +J/J-ar/il+c22-2I +3^-462*. 


CHAPTER  XXIII 
EQUATIONS 

206.  Definitions.  —  An  equation  is  a  statement  that  two 
expressions  are  equal  in  value. 

Thus,  A=%ab  is  an  equation.  So  are  A  =?rr2;  V  =\irr3;  and 
3x+4  =  10. 

The  part  to  the  left  of  the  equality  sign  is  called  the  first 
member  of  the  equation,  and  the  part  to  the  right,  the  second 
member. 

If  the  area  of  a  rectangle  is  36  sq.  ft.  and  the  altitude  is 
4  ft.,  we  have  36  =46,  where  6  stands  for  the  base.  Now  it  is 
easy  to  see  that  the  statement,  or  equation,  is  true  if,  and  only 
if,  6  =  9.  Such  an  equation  as  this,  where  the  letter  whose 
value  we  wish  to  find  has  a  certain  value,  is  called  a  conditional 
equation.  That  is,  this  equation  is  true  on  the  condition 
that  6  =  9,  and  for  no  other  value  of  6. 

Not  all  statements  of  equality  are  conditional.     For  instance, 

re2—  4 

.n  =x  —  2  is  an  equation;  but  x  may  have  any  value  what- 

ever, and  still  make  the  equation  true. 

9—4 
Thus,  if  £=3,  the  equation  becomes  =3—2  or  1=1.     If  a;  =  4 

~ 


we  get  2  =  2.     Similarly  for  any  value  we  give  x. 

This  kind  of  an  equation  is  called  an  identical  equation 
or  an  identity. 

The  number  asked  for  in  an  equation  or  the  letter  standing 
for  it  is  called  the  unknown  number,  the  unknown  quantity, 
or,  briefly,  the  unknown. 

The  following  definitions  are  stated  for  equations  involving 
one  unknown,  but  may  easily  be  extended. 

An  equation  that  is  true  only  on  condition  that  the  unknown 
has  particular  values  is  called  a  conditional  equation  or, 
briefly,  an  equation. 

17  257 


258  PRACTICAL  MATHEMATICS 

An  equation  that  either  involves  no  unknown,  or  that  is 
true  for  any  value  whatever  that  may  be  given  to  the  unknown, 
is  called  an  identical  equation  or,  briefly,  an  identity. 

To  solve  an  equation  is  to  find  the  value  or  values  of  the 
unknown  that  will  make  the  equation  true. 

207.  The   equation. — A   large   number  of   problems,   that 
are  solved  by  means  of  algebra,  involve  the  equation  in  one 
form  or  another.     This  makes  the  equation  the  most  important 
tool  of  algebra;  in  fact,  it  may  be  looked  upon  as  a  more  or 
less  complicated  piece  of  machinery,  with  which  the  student 
should  become  very  familiar. 

To  become  familiar  with  the  mechanism  of  the  equation 
and  its  applications  requires  a  great  deal  of  time  and  much 
drill  in  solving  equations.  Much  of  the  work  in  solving 
equations  is  mechanical,  in  that  it  does  not  require  much 
thought  in  its  performance.  However,  there  is  a  reason  for 
doing  each  step  that  is  taken,  and  one  should  be  able  to  give 
this  reason. 

208.  Solution  of  equations. — As  already  stated,  to  solve 
an  equation  is  to  determine  the  value  or  values  of  the  unknown 
number  or  numbers  in  the  equation.     This  may  be  an  easy 
matter,  but  it  is  often  difficult.     Here  we  shall  start  with  very 
simple  equations  and  endeavor  to  discover  certain  general 
methods  of  procedure  in  the  solution. 

Example  1.     Find  the  value  of  x,  if  x  —  5  =  3. 

Here  one  readily  sees  by  inspection  that  x  =  S,  but  this 
does  not  help  us  in  solving  a  more  complicated  equation. 
If,  however,  we  notice  that  in  order  to  determine  x  =  8,  5  is 
added  to  each  member  of  the  given  equation,  we  have  a  method 
of  procedure  that  we  can  apply  to  another  like  problem.  We 
have  then  the  solution: 
Given  equation,  x  — 5  =  3. 

Adding  5  to  each  member,  x  =  3+5. 

Collecting  the  terms,  x  =  8. 

Example  2.     Solve  for  x,  if  z+3  =  10. 

Solution.     Given  equation,  z+3  =  10. 
Subtracting  3  from  each  member,  x—  10  —  3. 
Collecting  the  terms,  x  =  7. 

Example  3.     Solve  for  6,  if      46  =  36. 


EQUATIONS  259 

Solution.     Given  equation,       46  =  36. 
Dividing  each  member  by  4,          6  =  9. 

Example  4.     Solve  for  a:,  if  4z + 5  —  7  =  2x + 6. 

Solution.     Given  equation,  4z+5  —  7  =  2z+6. 

Adding  7  to  both  members,  4z-f-5  =  2£+6+7. 

Subtracting  5  from  both  members,  4z  =  2z  +  6  +  7  —  5. 

Subtracting  2z  from  both  members,      4z  —  2z  =  6  +  7  —  5. 
Collecting  the  terms,  2x  =  8. 

Dividing  both  members  by  2,  a:  =  4. 

Notice  that  when  a  term  is  added  to  or  subtracted  from 
both  members  of  an  equation,  it  is  transposed  from  one  member 
to  the  other  and  its  sign  is  changed.  Now  by  this  transposing 
we  can  bring  all  the  terms  that  contain  the  unknown  into  the 
first  member  and  all  the  others  into  the  second  member.  This 
gives  a  convenient  form,  for  we  wish  finally  to  have  an  equa- 
tion in  which  the  form  is: 

unknown  =  some  number. 

Steps  in  solution.  The  solution  of  an  equation  that  is  in  a 
simple  form  may  then  be  carried  out  in  the  following  three 
steps : 

(1)  Transpose  all  terms  containing  the  unknown  to  the  first 
member,  and  all  other  terms  to  the  second  member.     In  each  case 
change  the  sign  of  the  term  transposed. 

(2)  Collect  the  terms  in  each  member. 

(3)  Divide  each  member  by  the  coefficient  of  the  unknown. 

It  will  be  found  later  that  there  are  other  changes  to  be  made 
in  an  equation  that  is  not  in  a  simple  form,  before  these  three 
steps  are  to  be  performed. 

209.  axioms. — An  axiom  is  a  truth  that  we  accept  without 
proof. 

The  solutions  of  the  equations  and  the  changes  mentioned 
in  the  preceding  article  suggest  the  following  axioms: 

(1)  //  equal  numbers  are  added  to  equal  numbers,  the  sums 
are  equal. 

(2)  //  equal  numbers  are  subtracted  from  equal  numbers,  the 
remainders  are  equal. 

(3)  //  equal  numbers  are  multiplied  by  equal  numbers,  the 
products  are  equal. 


2fiO  PRACTICAL  MATHEMATICS 

(4)  //  equal  numbers  are  divided  by  equal  number*,  the  quotients 
are  equal. 

(5)  Numbers  that  arc  equal  to  the  same  number  vr  e<juul 
numbers  are  equal  to  each  other. 

(6)  Like  powers  of  equal  numbers  are  equal. 

(7)  Like  roots  of  equal  numbers  are  equal. 

(8)  The  whole  of  anything  equals  the  sum  of  all  its  parts. 
210.  Testing  the  equation. — The  equation  puts  the  question : 

What  number,  if  any,  must  the  unknown  represent  in  order 
that  the  two  members  of  the  equation  shall  be  equal?  The 
solution  of  the  equation  answers  this  question,  but  it  is  always 
well  to  test  or  check  the  work.  This  may  be  done  by  substi- 
tuting the  number  obtained  for  the  unknown  in  place  of  the  un- 
known letter.  If  the  two  members  of  the  equation  then  be- 
come identical,  the  number  substituted  is  the  answer  to  the 
equation. 

Example.     Solve  and  test:    47r- 17  =235-37r. 

Solution.     Given  equation,    47r— 17  =235  — 37r. 

Transposing,  47r+37r  =  235+17. 

Collecting  terms,  84r  =  252. 

Dividing  by  the  coefficient  of  r,          r  =  3. 

Testing  by  substituting  3  for  r  in  the  equation, 

141-17  =  235-111. 

Collecting  gives  the  identical  equation,  124  =  124. 

EXERCISES  62 

Solve  and  test  the  following  equations: 

1.  3z+4=2z+5.  Ans.  1. 

2.  3z-4=z  +  12.  Ans.  8. 

3.  3z-25+2z=39-3z.  Ans.  8. 

4.  250z-20  =  20z+440.  Ans.  2. 

5.  5z+7=2z+9.  Ans.  I 

6.  4z-4=z+7.  Ans.  3\. 

7.  17z  +  10  =  14z  +  16.  Ans.  2. 

8.  3z  +  14+2z=z+26.  Ans.  3. 

9.  40z-10  =  15z+90.  Ans.  4. 

10.  8z+25  =  2z+28.  Ans.  I 

11.  16z-3=6z+8-23z.  Ans.  i 

12.  »-2+7y~14y+7-8y.  Ana.  4J. 

13.  2z  +  (5z-5)=7.  Ans.  If. 


EQUATIONS  261 

Suggestion.     First  clear  of  parentheses  and  then  proceed  as  before. 

14.  I2y-(2y  +  l)=38+7y.  Ans.  13. 

16.  (j/+3)-27  =  10-y-(y-2).  Ans.   12. 

16.  18y-(2y+6)-17  =  14y-(14-y).  Ans.  9. 

17.  35a-64  +  (13-a)  =  16a-(47-14o).  Ans.   1. 


211.  The  equation  in  solving  problems.  —  As  has  been  stated 
before,  one  of  the  things  to  be  gained  in  the  study  of  mathe- 
matics is  to  be  able  to  express  ideas  in  mathematical  symbols. 
The  student  has  already  had  some  practice  in  this.  We  have 
stated  various  principles  and  rules  as  formulas,  and  have  done 
much  translating  from  English  to  mathematical  language  and 
from  the  language  of  mathematics  to  English  in  Chapter  XX. 

For  this  translating  we  cannot  give  rules  as  we  can  for  the 
operations  to  be  performed  in  the  solutions  of  exercises.  The 
student  must  first  thoroughly  understand  the  thing  to  be 
expressed;  and,  secondly,  he  must  know  the  signs  and  symbols, 
that  is,  the  language  of  mathematics.  The  following  sugges- 
tions will  help  the  student  to  state  a  problem  in  the  form  of  an 
equation. 

(1)  Read  carefully  the  statement  of  the  problem,  as  it  is 
given  in  words. 

(2)  Select  the  unknown  number  and  represent  it  by  some 
letter.     If  there  are  more  unknown  numbers  than  one,  try  to 
express  the  others  in  terms  of  the  one  first  selected. 

(3)  Find  two  expressions  which,  according  to  the  problem, 
represent  the  same  number,  and  set  them  equal  to  each  other. 
This  forms  the  equation  to  be  solved. 

EXERCISES  63 

1.  A  man  is  m  years  old.     Give  in  algebraic  symbols  his  age  10  years 
ago.     Give  his  age  n  years  age.     Give  his  age  s  years  from  .now. 

2.  A  man  is  m  years  old.     When  will  he  be  50  years  old  r     When  was 
he  15  years  old?     When  a  years  old? 

3.  A  train  runs  n  miles  in  1  hour.     How  many  miles  will  it  run  in 
8  hours?     In  t  hours?     In  TO  minutes?  ^ns    gn-  nt-  mn  . 

4.  A  train  runs  x  miles  in  10  hours.     How  far  does  it  run  in  1  hour? 
In  n  hours?     How  long  will  it  take  to  run  100  miles? 

x    nx    1000 


_'»,_>  PRACTICAL  MATHEMATICS 

5.  How  many  cents  in  a  dollars?     In  6  dimes?     In  c  dollars  and  d 
dimes?  Ans.  lOOo;  106;  lOOc+lOd 

6.  If  x  represents  the  number  of  bushels  of  apples  bought,  what  was 
the  price  per  bushel  if  $10  was  the  cost  of  all? 

7.  Find  the  cost  per  bushel  of  apples  if  25  bushels  at  x  dollars  per 
bushel  cost  $10.     State  as  an  equation  and  solve. 

8.  A  room  is  three  times  as  long  as  it  is  wide  and  its  perimeter  is  96  ft. 
Find  the  length  and  width. 

Solution. 

(1)  Let  x  =  number  of  feet  in  width. 

(2)  Then  3x  =  number  of  feet  in  length, 

(3)  and  z+i+3z  +3z=  perimeter. 

(4)  Also  96=  peri  meter. 

(5)  /.  z+x+3z+3z  =  96,  by  axiom  (5). 

(6)  Collecting  terms,  8x  =  96. 

(7)  Dividing  by  8,  x  =  12  =  number  of  feet  in  width, 

and  3z  =  36  =  number  of  feet  in  the  length. 

Note  that  in  statements  (3)  and  (4)  are  two  expressions  for  the  same 
thing,  the  perimeter.  These  two  expressions  put  equal  to  each  other  in 
statement  (5)  give  the  equation  to  be  solved. 

9.  In  a  company  there  are  64  persons,  and  the  number  of  children  is 
three  times  the  number  of  adults.     How  many  are  there  of  each? 

Solution. 

(1)  Let  x  =  the  number  of  adults. 

(2)  Then  3x  =  the  number  of  children, 

(3)  and  z+3x  =  number  in  the  company. 

(4)  Also  64  =  number  in  the  company. 

(5)  .'.  z+3*  =  64. 

(6)  4x=64. 

(7)  x  =  16  =  the  number  of  the  adults. 

(8)  3x  =  48  =  the  number  of  the  children. 

10.  If  twice  a  number  is  added  to  six  times  the  number  the  sum  is  96. 
What  is  the  number?  Ans.  12. 

11.  A  horse  and  wagon  cost  together  $214.     If  the  horse  cost  $76  more 
than  the  wagon,  find  cost  of  each.  Ans.  horse  $145;  wagon  $69. 

12.  The  second  of  three  numbers  is  4  times  the  first  and  the  third  ie 
3  times  the  first.     If  the  first  and  second  are  added  and  the  third  sub- 
tracted from  the  sum,  the  remainder  is  60.     Find  the  three  numbers. 

Ans.  30;  120;  90. 

13.  The  number  of  copies  of  a  book  sold  doubled  each  year  for  three 
years,  and  in  that  time  36,750  copies  were  sold.     How  many  were  sold 
each  year?  Ans.  5250;  10,500;  21,000. 

14.  A  rectangular  lot  is  30  rods  longer  than  it  is  wide.     Use  w  rods  for 
the  width  and  state  in  an  equation  that  the  perimeter  is  260  rods.     Solve 
this  equation  and  find  tho  width  and  length  of  the  field. 

Ans.  50  rd. ;  80  rd. 


EQUATIONS  263 

15.  The  sum  of  two  numbers  is  300  and  their  difference  is  200.     What 
are  the  numbers? 

Solution. 

(1)  Let  x  =  the  greater  number. 

(2)  Then  300—  x  =  the  lesser  number, 

(3)  and  x- (300- x)  =  the  difference. 

(4)  Also  200=  the  difference. 

(5)  /.  x-(300-x)  =200,  by  axiom  (5). 

(6)  Simplifying,  x-   300+x  =  200. 

(7)  Transposing,  x  +  x  =  300 +200. 

(8)  Collecting  terms,  2x  =  500. 

(9)  Dividing  by  coefficient  of  x,  x=250,  the  greater  number. 

(10)  300-x=50,  the  lesser  number. 

Test.  The  sum  of  250  and  50  is  300,  and  the  difference  is  200.     Hence 
the  conditions  of  the  problem  are  satisfied. 

16.  A  man  has  $77  in  $1  bills  and  $10  bills.     How  many  bills  has  he 
if  he  has  the  same  number  of  each?  Ans.  14. 

Suggestion. 

Let  x  =  number  of  each  kind  of  bills. 

Then  x  =  number  of  dollars  represented  in  $1  bills, 

and  lOx  =  number  of  dollars  represented  in  $10  bills. 

.'.  x  +  10x=77. 

17.  If  I  have  three  times  as  many  $5  bills  as  $2  bills  and  the  amount 
of  money  in  these  bills  is  $85,  how  many  of  each  kind  of  bills  have  I? 

Ans.  15;  5. 

18.  In  my  pocketbook  are  a  certain  number  of  silver  dollars,  twice  as 
many  quarters  as  dollars,  and  five  times  as  many  dimes  as  dollars.     If 
the  total  amount  of  money  is  $4.00,  find  the  number  of  coins  of  each  kind. 

Ans.  2;  4;  10. 

19.  A  shopper  bought  three  articles.     The  second  cost  three  times  as 
much  as  the  first  and  the  third  $3  more  than  the  second.     Find  the  cost 
of  each  if  the  total  cost  was  $9.  Ans.  $§ ;  $2| ;  $5f . 

20.  Three  tanks  hold  a  total  of  24,500  gallons.     The  first  holds  4500 
gallons  more  than  the  second,  and  the  second  2500  gallons  more  than 
the  third.     How  many  gallons  does  each  hold? 

Ans.  12,000;  7500;  5000. 

21.  Of  two  candidates  for  the  same  office,  the  successful  one  received 
a  majority  of  265.     How  many  votes  did  each  receive,  if  the  total  number 
of  votes  cast  was  6793V  Ans.  3529;  3264. 

22.  Find  the  three  consecutive  even  numbers  whose  sum  is  216. 

Ans.  70;  72;  74. 
Suggestion. 
Let  x  =  first  number. 
Then  x+2=second  number,  and  x+4=third  number. 

23.  Find  the  four  consecutive  odd  numbers  whose  sum  is  88. 

Ans.  19;  21;  23;  25. 


264  PRACTICAL  MATHEMATICS 

24.  Find  the  three  consecutive  numbers  whose  sum  is  66. 

Ans.  21;  22;  23. 

26.  Divide  $210  between  A,  B,  and  C  so  that  B  shall  have  $35  less  than 
A  and  $20  more  than  C.  Ans.  $100;  $65;  $45. 

26.  One  angle  is  the  complement  of  another.     If  14°  is  subtracted  from 
the  second  and  14°  added  to  the  first,  the  first  will  be  44°  larger  than  the 
second.      Find  the  two  angles.  Ans.  53°  and  37°. 

Two  angles  are  said  to  be  complements  of  each  other  if  their  sum  is 
90°.     (See  Art.  90.) 

27.  The  difference  between  two  angles  is  14°.     Find  the  angles  if  they 
are  complements  of  each  other.  Ans.  52°  and  38°. 

28.  A  father  and  son  earn  $188  a  month.     If  the  son's  wages  were 
doubled,  he  would  receive  $62  less  than  his  father.     How  much  does  the 
son  receive?  Ans.  $42. 

29.  Three  men,  A,  B,  C,  raised  4080  bushels  of  wheat.     A  raised  three 
times  as  many  bushels  as  B  and  330  bushels  more  than  C.     How  many- 
bushels  did  each  raise?          Ans.  A,  1890  bu. ;  B,  630  bu. ;  C,  1560  bu. 

30.  The  sum  of  four  angles  about  a  point  is  360°.     The  second  is  twice 
the  first,  the  third  three  times  the  second,  and  the  fourth  is  10°  greater 
than  the  first.     Find  the  angles.  Ans.  35°,  70°,  210°,  45°. 


CHAPTER  XXIV 
MULTIPLICATION 

212.  Fundamental  ideas. — Multiplication  of  whole  num- 
bers in  arithmetic  may  be  thought  of  as  a  shortened  process  of 
addition.  For  instance, 

5X3  =  5+5  +  5  =  15.  (1) 

We  say  that  the  multiplicand  is  used  as  an  addend1  as 
many  times  as  there  are  units  in  the  multiplier. 

This  idea  of  multiplication  must  be  enlarged  in  order  to 

8X3 
include  the  multiplying  by  a  fraction.     Thus,  8Xf=     .     ', 

that  is,  we  multiply  by  the  numerator  and  divide  by  the  denomi- 
nator of  the  multiplier. 

For  multiplication  in  algebra  we  shall  retain  its  arithmetical 
meaning  when  the  multiplier  is  a  positive  whole  number  or 
fraction,  but  shall  have  to  extend  the  meaning  to  include 
negative  numbers. 

From  the  arithmetical  meaning,  since  the  multiplier  is 
positive,  we  have, 

(-5)X(+3)  =  (-5)  +  (-5)  +  (-5)  =  -15.  (2) 

Now  we  know  that  when  we  multiply  two  positive  abstract 
numbers  together,  it  does  not  matter  which  is  used  as  the 
multiplier.  If  we  assume  that  this  principle  holds  when  one 
of  the  numbers  is  negative,  we  have, 

(-5)X(+b)  =  (+3)X(-5)  =  -15. 

This  gives  us  the  following  meaning  for  multiplication  by  a 
negative  number: 

To  multiply  by  a  negative  number  is  to  multiply  by  its  abso- 
lute value  and  then  change  the  sign  of  the  product. 

Thus,   to  multiply   +5  by    —3   we  multiply   +5  by   +3 

getting  +15,  and  then  change  the  sign  of  the  result.     That  is, 

(+5)X(-3)=  -15.  (3) 

1  An  addend  is  one  of  the  numbers  to  be  added  in  an  addition  problem. 

265 


PRACTICAL  MATHEMATICS 

Likewise,  to  multiply  —5  by  —3  we  multiply  —5  by  +3 
getting  —  15,  and  then  change  the  sign  of  the  result.  That  is, 

(-5)X(-3)  =  +  15.  (4) 

In  (1),  (2),  (3),  and  (4)  we  have  examples  of  all  the  combina- 
tions possible  of  two  algebraic  numbers. 

213.  From  the  above  considerations  we  see  that  in  finding 
the  product  of  two  algebraic  numbers: 

(1)  The  numerical  part  of  the  product  is  the  product  of  the 
absolute  values  of  the  multiplicand  and  multiplier. 

(2)  The  sign  of  the  product  is  plus  when  the  signs  of  the 
multiplicand  and  multiplier  are  alike,  and  minus  when  their 
signs  are  unlike. 

This  is  called  the  law  of  signs  in  multiplication.  It  may 
be  stated  as  follows: 


+  x-  =  -, 
-x  +  =  -. 

214.  Concrete  illustration.  —  For  those  who  find  the  fore- 
going difficult  to  understand,  the  following  may  clear  up 
matters. 

There  is  a  machine  shop  employing  laborers  and  appren- 
tices. The  laborers  are  paid  $15  per  week,  and  the  apprentices 
are  charged  $3  per  week. 

Suppose  that  an  increase  in  the  number  of  men  or  dollars  is 
positive,  and  a  decrease  in  either  is  negative.  Thus,  a  number 
of  laborers  or  apprentices  taken  in  will  be  called  positive,  and 
a  number  let  go  will  be  called  negative;  while  the  number  of 
dollars  received  from  an  apprentice  is  positive  and  the  number 
of  dollars  paid  a  laborer  is  negative. 

On  these  suppositions  we  have: 

(1)  If  apprentices  are  increased  by  5,  the  amount  of  money 
is  increased  $15  per  week.  That  is, 


(2)  If  apprentices  are  decreased  by  5,  the  amount  of  money 


is  decreased  $15  per  week.     That  is, 


MULTIPLICATION  267 

(3)  If  laborers  are  increased  by  5,  the  amount  of  money  is 
decreased  by  $75  per  week.     That  is, 

(-15)X(+5)  =  -75. 

(4)  If  laborers  are  decreased  by  5,  the  amount  of  money  is 
increased  by  $75  per  week.     That  is, 

(-15)X(-5)  =  +75. 

From  these  considerations,  we  may  deduce  the  same  rules 
as  already  given. 

215.  Continued  products.  —  To  find  the  product  of  three 
or  more  numbers,  we  find  the  product  of  the  first  two,  and 
then  multiply  this  product  by  the  third,  and  so  on  till  all 
the  numbers  have  been  used. 

By  applying  principles  (1)  and  (2)  of  Art.  213,  we  obtain  the 
following  : 

(1)  The   product  of  an   odd  number  of  negative  factors   is 
negative. 

(2)  The  product  of  an  even  number  of  negative  factors   is 
positive. 

(3)  The  product  of  any  number  of  positive  factors  is  positive. 

Thus,     (-2)  (-2)  (-2)  (-2)  (-2)  =  -32,  while 
(-2X-2)(-2X-2)(-2)(-2)  = 


The  first  one  of  these  equals  (  —  2)5,  and  is  then  read  "the 
fifth  power  of  —2."  The  second  is  (  —  2)6. 

It  should  be  noted  that  such  a  form  as  (  —  2)5  does  not 
mean  the  same  as  —  25,  though  they  may  be  equal.  The 
form  —  25  is  read  "minus  2  to  the  fifth  power.1' 

Thus,     (-2)2  =  ^-2X-2)=4  =  22, 

(-2)»  =  (-2K-2)(-2)  =  -8=  -23, 


-2«=-  (2)  (2)  (2)  (2)  (2)  =  -32, 

(-3)*(-2)3  =  (-3X-3)(-2)(-2)(-2)  =  -72, 
(42X3')=  4-4-3-3  =  144. 

EXERCISES  64 

1.  Find  the  product  of  -7,  -8,  and  +10.  4ns.  560. 

2.  Find  the  product  of  -2,  -40,  +75,  and  -60. 

3.  Find  cube  of  -7,  of  +8,  of  -21. 


•Jl.s  PRACTICAL  MATHEMATICS 

Find  values  of  the  following: 

4.  (-4)*.      Ans.   -1024.  5.  (-8)4.  Ana.  4096. 

6.  (-1)">.     Ana.   +\.  7.  (-4)«(-3)».         Ans.   -576. 

8.  2(+6)*(-6)».  Ana.  2592. 
If  z  =  2,  j/=  —3,  z  =  —4,  find  value  of: 

9.  *V-       4rw.  72.  10.  xyz.  Ana.  24. 
11.  j/'z.         An«.   108.                      12.  zV«f-                           Ana.  576. 
IS.  (6)(-5)(-2)t/>.                                                            ATM.    -1620. 

14.  15-(-2)-(-3)+4-(-5)-6.  Ana.    -30. 

15.  l-2-3+3-4-5  +  (-5)-(-6)-(-7).  Ana.   -144. 

16.  4-(-3)-5+2-(-3)-4+3-(-3)-4.  Ans.   -120. 

17.  (-2)»-(-3)«-(-4)-6-(-7).  Ans.   -2760. 

18.  (-3)J-3»  +  (-2)s-2s.  4rw.   -118. 

19.  (-2)-(-3)  +  (-4)-(-5)  +  (-5)-(-6).  Ans.  56. 

20.  l-2-3-4-5-(-l)(-2)(-3)(-4)(-5).  Ans.  240. 
31.  4-5-6  7-(-4)(-5)(-6)(-7).  Ans.  0. 

216.  Law  of  exponents.  —  The  law  which  applies  to  ex- 
ponents that  are  positive  integers  is  derived  from  the  defini- 
tion given  in  Art.  186. 

Since  a6  =  a-a-a-a-o, 
and     a3  =  a-a-a, 
then    o6-a3  =  a-a-a-a-a-a-a-o  =  a8, 
and     a5-a3  =  a6+3  =  a8. 

In  general  an  =  a-a-a-o--  to  n  factors1 
and     am  =  a-a-a-a—  to  ra  factors, 
then    an-am  =  a-a-a-a-a--'  to  (n+m)  factors, 
and     an'(im  =  an+m. 

Similarly,  when  there  are  any  number  of  factors  we  have 


LAW.  The  product  of  two  or  more  powers  of  the  same  base 
is  equal  to  that  base  affected  with  an  exponent  equal  to  the  sum 
of  the  exponents  of  the  powers. 

217.  To  multiply  a  monomial  by  a  monomial.— 

Example.     Multiply  14a362  by  —  3a463. 

Process. 

Discussion.     Since  the  multiplier  is  composed  of        14a362 
the  factors  —3,  a4,  and  63,  the  multiplicand  may  be 
multiplied  by  each  successively.     In  each  case  the 
product  for  any  one  of  these  factors  is  obtained  by  multi- 

1  A  repetition  of  dots,  as  a,  b,  c,--.  is  the  sign  of  continuation.  It  is 
read  "and  so  on." 


MULTIPLICATION  269 

plying  a  single  factor  in  the  multiplicand  by  it.  We 
multiply  by  —3,  by  multiplying  14  by  —3,  which  gives 
-42a362.  This  is  multiplied  by  a4,  by  multiplying  the  a3 
by  a4,  which  gives  —  42a762.  This  is  multiplied  by  b3,  by 
multiplying  the  62  by  b3,  which  gives  —  42a765,  the  answer. 

The  multiplication  is  carried  out  by  determining  in  the 
following  order: 

(1)  the  sign  of  the  product, 

(2)  the  coefficient  of  the  product, 

(3)  the  letters  of  the  product, 

(4)  the  exponents  of  these  letters. 

Thus,  in  the  above  example  the  sign  is  +  X  —  =  —  ;  the  coefficient  is 
14X3  =  42;  the  letters  are  a  and  b;  and  the  exponents  are,  for  a,  3+4  =  7, 
and  for  b,  2+3=5. 

This  plan  should  be  carefully  followed  by  the  beginner. 
218.  To  multiply  a  polynomial  by  a  monomial.  — 

RULE.  The  product  is  found  by  multiplying  each  term  of 
the  multiplicand  by  the  multiplier,  and  taking  the  algebraic 
sum  of  these  partial  products. 

Example.     Multiply  7  ax3  -  21a64  -  3x2  by  2a263z4. 

Process.     7ax3-21ab*-3x2 


-  42a36  V  - 

Explanation.  The  first  term  at  the  left  of  the  product 
is  obtained  by  multiplying  the  first  term  at  the  left  of  the 
multiplicand  by  the  multiplier.  The  second  and  third  terms 
in  the  product  are  obtained  in  a  similar  manner.  In  each  of 
the  multiplications  we  have  a  monomial  by  a  monomial, 
which  has  been  discussed  in  the  previous  article. 

219.  To  multiply  a  polynomial  by  a  polynomial.  — 

RULE.  Multiply  every  term  of  the  multiplicand  by  each  term 
of  the  multiplier,  write  the  like  terms  of  the  partial  products  under 
each  other,  and  find  the  algebraic  sum  of  the  partial  products. 

Example  I  .     Multiply  x2  -\-  3xy  —  2y  2  by  2xy  -2y*. 

Process. 


2xy-2yz 


2xy  times  (z2+3z?/  —  2?/2)  =         2x3y+Qx*yz  —  4xy3 
-2y2  times  (z2  +  ?>xy  -  2?/2)  =  -2a?V- 

Adding  these  we  get,  2x3y+4:x'2y2  — 


270  PRACTICAL  MATHEMATICS 

Example  2.     Multiply  3a2+362+o&  by  &»-2al6+a&2. 

Process. 
3a2+362    +06 
6»-2a2b  +ab« 

3a2&3+366+a&4 

-6o26s  -6a46-2a»62 

a2b3          +3afr4 


-2a2&3+365+4a&4-6a4&+  a362 

220.  Test.  —  Problems  in  multiplication  can  be  tested  by 
substituting  convenient  numerical  values  for  the  letters. 
It  is  best  to  use  values  larger  than  1,  since  with  1  the  exponents 
are  not  tested,  as  any  power  of  1  is  1. 

Test  of  example  1,  by  letting  x  —  1  and     y  =  2. 

z2+3xy-27/2  =4+12-8  =  8 

2zy-2y*  =8-8         =0 


2x*y  +  4s22/2  -  lOxy3  +  4t/4  =  32  +  64  -  1  60  +  64  =  0. 
The  work  is  probably  correct  if  the  product  of  the  values 
of  the  two  factors  equals  the  value  of  the  product. 

EXERCISES  65 

Find  the  product  of  the  following: 

1.  10o62  and  3o'6.  Ans.  30a46». 

2.  16n'x5  and  -2n'xt. 

3.  4a2xV  and  —  5x*y.  Ans.    -20a*xV- 

4.  -17iV&nd  -3a'x2. 

5.  -5m»n2<iV  and  -2mlon«cV-  Ans.   10mlin*c*d*y>. 

6.  —  x*y*  and  —x*y*z*. 

7.  —  3x*ys,  x2,  t/3,  and  4xy.  Ans.   —  12x*j/T. 

8.  3PQ,  Q«,  and  4P2Q». 

9.  —  §ox2,  Jo2xs,  and  — 

10.  a",  a2B+1,  o8"+I,  and  o"+1.  Ans.  a"n+*. 

11.  (a3)2  or  a3  times  a*.  Ans.  a*. 

12.  (m2*!3)^?  ATM.  m"n12. 

13.  (4a26 

14.  (an6")'«?  Ans. 

15.  as62c 
18.  x4j/2z 

17.  3<z4c2--4a6c--2<i«c<.  Arw.  24aI*cT. 

18.  p^r2  —  pV"  —  p^V. 

19.  (6xj/z)-(-7axj/2)-(-2o3xVz*). 

20.  562-(5+662-764).  Ans.  2562+30&4-35&«. 


MULTIPLICATION 


271 


21.  -7x2y(.-3x22/2+2x?/). 

22.  25a62-(2x463+2c2d). 

23.  -5x-(-3a+2a2+4). 

24.  -15x72/8-(m7n8x2?/2-13pV0)-    Ans.  - 


Ans.  21x*y3-Ux3y2. 
Ans.  50a66x4  +50a62c2d. 
Ans.  15ax-10a2x-20x. 


OK       Q-r 3i*5;f7.  ^8^/6^4 1  Q  H?/2^ 

av»       — Oit/    y    6       \Jir    if    6  .LOU/    y    )» 

26. 
27. 

28.  2x-3yby  4x+y. 

29.  x2-2y  by  4x+y2. 

30.  -_3xJ;7  by  2x-l. 
,31.  5x+4?/by  3x-2y. 

32.  ax— by  by  ax+6y. 

33.  a2x2-2y2  by  a2x2+2y2. 

34.  a2+a6+62by  a -6. 

35.  a2-a6+62  by  a+6. 

36.  tf-ab+b2  by  2a-46. 

37.  2x+3y-z  by  2x-3y+z. 

38.  a2+a6+62  by  a2-a6+62. 

39.  x2-2xy+y"-  by  x2+2xy+i 

40.  6x2+2x  +  l  by  x2— x-1. 

42.  x2-4x  +  16  by  x+5. 


Ans.   - 
Ans. 
Ans.    - 

Ans.  8x2-10x7/-3i/2.-/ 

Ans.  4x3—  8xy+x2y2  —  2y 

Ans.    - 

Ans.  15x2+2xy-8y*. 

Ans.  a2x2-bzy2. 

Ans.  a4x4  — 4y4. 

Ans.  a3  —  b3- 

Ans.  a3+63. 

Ans.  2o3  -  6a26  +  6a62  -  463. 
Ans.  4x2 

Ans. 

Ans.  a:4—  2x2y2+ y4. 
Ans.  6x4-4x3-7x2-3x-l. 
Ans.  c4  - 

Ans. 
Ans.  n3-48n2-200n-200. 


43.  n2-50n-100by  n+2. 

44.  a3-3a26+3a62-&3by  a2-ab. 

Ans.  as-4atb+Qa3b2-4a2b3+ab4. 
46.  4x3—  3x2y+5xy2  —  6t/3  by  5x+6t/. 

Ans.  20x4+9x32/+7xV-36?/4. 

46.  4y2-W+2y  by  2j/2-3y+5.          Ans.  8y*-Gy2-8y3  +40^-50. 

47.  a4-a36+a262-a63+64by  a+b.  Ans.  a3+65. 

48.  2ac2—  3by  by  2c3—  3y2.  Ans.    4ac5  —  Qbc3y— Qac2y?+Qby3. 

49.  o3+3a2x+3ax2+x3  by  a3-3a2x+3ax2-x3. 

Ans.  a6-3a4x2+3a2x4-x6. 

60    cc^  ~f~x^?/^  ~\~ij^  bv  x^ ?y*^  ^4.  TIS    3/^  —  v^ 

Signs  of  grouping  are  often  used  to  enclose  factors  of  a  product.  Thus, 
(a+b) (a  —  6)  means  the  same  as  (a+b)  times  (a  — 6).  To  free  the  ex- 
pression of  these  signs  of  grouping,  the  indicated  multiplications  are 
performed. 

Free  the  following  of  signs  of  grouping  and  simplify: 

61.  (0+6)  (0-6).  Ans.  o2-62. 

62.  (2x+7)(3z2-8).  Ans.  6x3+21x2-16x-56. 

63.  (3x2+ab)(3x2-a&).  Ans.  9x4-a262. 

64.  5(x2-a6)+6(x2+a6).  Ans.  Ilx2+o6. 

65.  -3a6(a6-a2)+2a6(a2-62).  Ans.    -3a262+5a36-2a63. 

66.  -2(-2a4+3a36-64)+3(o4+2a36-264).  Ans.  7a<-464. 

67.  2(x2-3x  +  l)  — (x+4)(x-l).  Ans.  x2  — 9x+6. 


68. 


Ans.  6n3  +  12n2-14n-4. 


272 


PR  A  CTICAL  MA  THEM  A  TICS 


Solution.     2(n  +  l)(n-l)(n+2)«     2n»+  4n»-  2n-4 
Subtracting,  4n(l-n)(n+3)      •  -4n»-  8n«+12n 


59. 


6n»  +  12n*-14n-4. 

y')(x*-y«+zi,'). 
Ana.    -3x4+7xty-2x^-xyt+y*+xtyt+xtyt-xy*. 


—  2S+6— 

*  -3-2-  -> 

T 


FIG.   193. 


FIG.  194. 


60.  find  the  area  of  a  rectangle  2x+5  ft.  in  length  and  x—  6  ft.  wide. 

Ana.  2x2-7x-30sq.  ft. 

61.  Find  the  volume  of  a  rectangular  solid  2z—  3  ft.  wide,  7x  —  2  ft. 
long,  and  x  +4  ft.  deep.  Ann.  14x3+31z*—  94x+24  cu.  ft. 

62.  Find  the  volume  of  a  right  circular  cylinder  if  the  altitude  is  h  ft. 
and  the  radius  2A-4  ft.  Ana.  ir(4A3-16/i2  +  16A)  cu.  ft. 


1 


Area  = 

ab 

FIG.  195. 


(1) 

ac 

(2) 
be 

f 

f 
1 

t<   a   >|<              b             >] 

FIG.   196. 

63.  A  hollow  square  has  dimensions  as  shown  in  Fig.  193.     Find  its 
area.  Ans.  3s*+28s+32. 

64.  A  ring  has  dimensions  as  shown  in   Fig.    194.     Find  its  area. 

Ans.   M27a*  +  150a+32). 

221.  Representation  of  products.  —  If  a  is  the  number  of 
units  in  the  altitude  in  Fig.  195  and  6  is  the  number  of  units  in 
the  base,  then  the  product  ab  is  the  number 
of  square  units  in  the  area  of  the  rectangle 
in  the  figure. 

Similarly,  the  two  rectangles  in  Fig.  190 
represent  the  product  of  (a+6)  by  c.     The 
part  marked  (1)  represents  the  partial  pro- 
duct ac,  and  the  part  marked  (2)  repre- 
sents the  partial  product  be. 
Fig.    1<)7     shows     that     (x+T/Xz+y)  =x*+2xy+y-.     The 
whole  figure  is  a  square  x+y  on  a  side.     This  is  made  up 
of  a  square  x  units  on  a  side,  having  x~  square  units;  a  square  y 


Fio.  197. 


MULTIPLICATION  273 

units  on  a  side,  having  y2  square  units;  and  two  rectangles 
each  x  units  by  y  units,  having  xy  square  units  each.  Hence 
the  whole  figure  contains  x2 -\-2xy +?/2  square  units. 

EXERCISES  66 

Find  the  product  in  each  of  the  following,  and  show  the  product  and  the 
partial  products  by  drawings. 

1.   (a+b+c)d.  2.  (a+b+c)(d+e). 

3.   (a+6+c)2.  4.  (x+y)(x-y). 

5.  (z-2/)2.  6.   (a+b+c)(d+e+f). 

222.  Approximate  products. — By  multiplication  we  get  the 
formula  (l+a)(l+fe)  =  l  +  a+6+a&. 

If  in  this  formula  a  and  6  are  small  fractions,  the  product 
ab  will  be  very  small  compared  with  a  and  6.  The  value  of 
(l+a)(l+6)  will  then  be  approximately  l+a+6. 

Thus,  if  a  =  0.05  and  6=0.02,  the  approximate  value  of  (l+a)(l+6)  = 
1+0.05+0.02  =  1.07,  while  the  exact  value  is  1.05X1.02  =  1.071.  1.071 
-1.07  =  0.001  =  difference. 

Example.  If  a  =  0.01  and  6  =  0.02  find  the  per  cent  of 
error  in  the  product  (1 +a)  (1  +6)  if  the  term  ab  is  disregarded. 

Solution.  The  approximate  value  of  (1 +0.01)  (1 +0.02)  = 
1  +  0.01  +  0.02=1.03. 

The  exact  value  =  1+0.01 +0.02+0.0002=  1.0302. 

The  error  =  1.0302 -1.03  =  0.0002. 

The  per  cent  of  error  is  obtained  by  finding  what  per  cent 
0.0002  is  of  1.0302. 

/ .  0.0002  -T-  1 .0302  =  0.0002  -  =  0.02  -  %.     Ans. 

EXERCISES  67 

1.  Find  per  cent  of  error  if  term  ab  is  disregarded  when  a  =  0.001  and 
6=0.002.  Ans.  0.0002-%. 

2.  Calculate  the  value  of  1.002X1.05  to  three  decimal  places  by  using 
the  approximate  method.  Ans.   1.052. 

3.  To  how  many  places  can  the  product  of  1.02X1.0024  be  found  by 
the  approximate  method?  Ans.  4. 

4.  How  much  error  per  cent  is  there  in  assuming  (l+a)(l+6)  =l+a+& 
when  a  =0.003  and  6  =  0.005?  Ans.  0.0015-%. 

5.  In  assuming  (l+a)3  =  l+3a,  what  is  the  per  cent  of  error  when 
a  =0.0002?     When   a  =0.002?     When   a  =  0.02?     When   a  =  0.2? 

Ans.  0.000012-%;  0.0012-%;  0.114-%;  7.41-%. 

18 


274  PRACTICAL  MATHEMATICS 

6.  In  measuring  the  radius  of  a  circle  of  correct  length  1  ft.,  there  is  an 
error  of  1%.     Find  the  per  cent  of  error  in  the  area  of  the  circle  deter- 
mined from  the  measured  radius.  Am.  2%  nearly. 

7.  If  in  determining  the  area  of  a  triangle  by  drawing  to  scale  and 
measuring  the  base  and  altitude,  the  base  is  measured  \\%  and  the 
altitude  2%  too  large  respectively,  find  per  cent  of  error  in  area,  disre- 
garding the  product  term.  Am.  3J%. 

8.  Find  the  approximate  products  of  the  following: 
(1)1.003X1.012.  (4)0.97X0.98. 
(2)1.02X0.97.                                 (5)0.996X0.997. 
(3)  1.004X0.998.                             (6)  0.985X0.996. 

9.  In  measuring  a  rectangle  the  length  is  measured  2%  too  large  and 
the  width  3%  too  small.     If  the  area  is  computed  from  these  measure- 
ments, find  the  per  cent  of  error  in  the  area. 

Ans.  1%  nearly,  too  small. 

10.  If,  in  measuring  a  triangle,  the  base  is  measured  1  %  too  large  and 
the   altitude    1J%    too  small,   find  the  per  cent  of  error  in  the  area 
computed  from  these  measurements.  Ans.  \%  nearly,  too  large. 

11.  If  the  error  in  a  number  is  1%,  what  is  the  per  cent  of  error  in  the 
square  of  the  number?     In  the  cube  of  the  number? 

Am.  2%;  3%  nearly. 


CHAPTER  XXV 
DIVISION,  SPECIAL  PRODUCTS,  AND  FACTORS 

223.  Division. — Division  is   the  inverse  of  multiplication. 
That  is,  the  quotient  must  be  an  expression  that  multiplied 
by  the  divisor  will  give  the  dividend. 

From  the  law  of  signs  and  the  law  of  exponents  in  multi- 
plication we  have  the  following : 

(1)  In  dividing,  like  signs  give  a  positive,  and  unlike  signs 
give  a  negative  sign  for  the  quotient. 

(2)  In  dividing  powers  of  the  same  base,  the  exponent  of  the 
quotient  equals  the  exponent  of  the  dividend  minus  the  exponent 
of  the  divisor. 

This  applies,  by  definition  of  the  positive  integral  exponent, 
only  when  the  exponent  of  the  dividend  is  larger  than  the 
exponent  of  the  divisor. 

Thus,  a5-f-a3  =  o2;  but  when  the  exponents  are  equal  and  we  subtract 
we  obtain  the  exponent  0,  which  is  meaningless. 

In  dividing  the  same  powers  of  the  same  base,  the  quotient 
is  1. 

Thus,  a3-r-a3  =  l  and,  in  general,  an-5-an  =  l. 

224.  Division   of   one   monomial   by   another. — It  is  well 
for  the  beginner  to  carry  out  the  work  of  a  division  in  a  regular 
order    as  in  multiplication.     (See  Art.   217.)     The  steps  in 
division  are: 

(1)  Determine  the  sign  of  the  quotient. 

(2)  Determine  the  coefficient. 

(3)  Determine  letters  and  exponents. 

Remember  that  in  the  process  of  division  we  divide  where 
we  multiply  in  multiplication,  and  we  subtract  exponents 
where  we  add  exponents  in  multiplication. 

Example.  Divide  25a4#5  by  —  5a2x*. 

Process,  carried  out  in  steps. 

275 


276  PRACTICAL  MATHEMATICS 

25  -^  -5=  -5, 
a4-i-a2  =  a2, 

<r5_i_  /j.3  _  r2 

.1        .    •</     ^^  «v    , 

.'..25o4x5-^  —  5a2z3  =  -5a2z2. 

The  last  only  should  be  written  down  in  performing  the 
work.  The  first  three  steps  are  mental  operations  and  are 
placed  here  for  guidance. 

The  division  of  one  monomial  by  another  may  also  be  per- 
formed as  a  cancellation.  If  we  recall  that  an  expression 
like  4a263  means  4a-a-b-b'b,  we  may  write  16a365c8-i-4a263  in 
the  form. 


Now  cancel  the  factors  common  to  the  dividend  and  divisor. 
The  product  of  the  factors  remaining  in  the  dividend  is  the 
quotient. 

This  process  is  too  long  for  rapid  work,  but  it  may  clear 
up  points  in  division  that  trouble  the  student. 

225.  Test.  —  The  work  in  division  can  be  tested  the  same  as 
in  multiplication  by  substituting  convenient  values  for  the 
letters.  It  may  also  be  tested  by  multiplying  the  divisor  by 
the  quotient,  when  the  product  will  be  the  dividend. 

EXERCISES  68 

Divide  the  following  and  test  : 

1.  IGx'y  by  2xy.  3.  10x4yzs  by  3x»z». 

2.  14x»z4  by  -7xz.  4.    -22a»6  by  -2ab. 

6.  100a4x»  by  -5a4x. 

_«.  18a»6»xby  -3aJ6z.  Ans.   -606. 

7.  -20*V  by  -5xV-  Aiis.  4xi/J. 

8.  -42as7nV  by  7a*m2.  Ana.    —  6a«/3. 

9.  -13aVby  -13ar/J.  Ans.  a. 

10.  SaVby  -8aV-  Ans.   -1. 

11.  18a"64by  -6a'6.  Ana.   -3a»64. 

12.  3ax»  by  Tax.  Ana.  fx*. 

13.  5a4x3c  by  -2a*.  Ans.    -1a»xjc. 

14.  3s  24a*  by  3*-2*a*.  Ans.  3-2*. 

15.  -54a'c»by  -5*ac4.  Ans.  5*«lc. 

16.  7'ac4xj/*  by  -74oc4y.  Ans.    —Ixy. 


DIVISION,  SPECIAL  PRODUCTS,  AND  FACTORS     '277 

226.  Division  of  a  polynomial  by  a  monomial. — 

Example.    Divide  24a5?/3  —  96a5?/6  by  8a4t/3. 
Process.        8a^3)24a57/3-96a5y6 
3a      -12m/3  ' 

RULE.  Tfte  division  is  performed  by  dividing  each  term  of  the 
dividend  by  the  divisor,  beginning  at  the  left. 

EXERCISES  69 

Divide  the  following  and  test  by  multiplication: 

1.  Uax+2Say+84az  by  14a.  Ans.  x+2y+8z.' 

2.  12o3+3a4  +  18a5by  3a3.  Ans.  4+a+6a2/ 

3.  3x5-16x3  +  14x2  by  x2.  Ans.  3z3-16x  +  14. 

4.  2-33+4-32+34by  32.  Ans.  2-3+4 +32. 
6.  2-32-5-3-33-7-5by  32.                                          Ans.  2-5 -3-3-7-5. 

6.  25ax2?/3-10xy-5xVby  5x2y3.  Ans.  5a-2xy-l. 

7.  21a3x3— 7a2z2  +  14ax  by  Tax.  Ans.  3a2x2-ax+2. 

8.  42a3-lla2+28o  by  7a.  Ans.  6a2-lfa+4. 

9.  24xV-8x4y5-24x?/2by  8x.  Ans.   3xy2-x3ys-3y2./ 
10.  4xV+8xV-12x6?/4by  2x2y2.  Ans.  2x2 

Tl.  ia562+5a364-7a463by  -Ho262.          Ans.    -ia3- 

12.  3.25a7-5.2a6+9.75a3by  0.25a3.  Aws.   13a4-20.8a3+39. 

13.  a2(a+fe)+a3(a+6)  by  a2.  Ans.  a+&+a(a+&): 

14.  3(a-6)+6a(a-6)  by  (o-6y.  Ans.  3+6a. 
16.  (a-b)(c+d)+(a-b)(x+y)  by  (a-6).               Ans.  c+d+x+y.  • 

227.  Factors  of   a  polynomial  when  one  factor   is   a   mo- 
nomial.— In  exercise  1  of  the  preceding  list,   each  term  of 
14a#+28cw/+84az  can  be  divided  by  14a.     The  quotient  is 
x+2y+Qz.     Now    the    product    of    x+2y+Qz    and    14a    is 
14ax+28ay+84az.     We  say  that  14a  and  x+2y+Qz  are  the 
factors  of  14az+28a7/+84a2. 

The  factors  of  a  polynomial  similar  to  the  above  are  a  mono- 
mial, containing  all  that  is  common  to  each  term  of  the  poly- 
nomial, and  the  quotient  found  by  dividing  the  polynomial  by 
the  monomial. 

Example.     Factor  4o2z  — 2az2+6a2z2. 

The  monomial  factor  is  2ax.  This  is  seen  by  inspection. 
Dividing  the  polynomial  by  2ax  we  find  2a  —  x -\-3ax  which 
is  the  other  factor.  The  factors  are  written  in  the  form 
2ax  (2a  —  x + 3a#) . 

Thp  work  may  be  tested  by  finding  the  product  of  the  factors, 
which  gives  the  expression  that  was  to  be  factored. 


278  PRACTICAL  MATHEMATICS 

EXERCISES  70 

Find  factors  of  the  following  and  test  the  work  by  multiplication : 

1.  ay+ax+ac.  5.  zj— z'+z4. 

2.  14+21o.  6.  64a4+63a7. 

3.  20a +306+40.  7.  a46»-aJ64-aJ6". 

4.  2*3+2'-5.  8.  16+32a-24a». 

9.  (z-2)a  +  (z-2)6.  Ana.  (z-2)(a+6). 

Suggestion.  Consider  (x  —  2)a  and  (x—  2)6  as  the  terras  of  the  poly- 
nomial. Then  Or— 2)  is  common  to  the  two  terms,  and  is  the  factor  to 
use  as  a  divisor  in  finding  the  other  factor. 

10.  (c-d)64-(c-d)c*z.  Ans.  (c-d)(64-c*z). 

11.  (3a-5)n  +  (3a-5)p.  Ans.  (3a-£)(n+p). 

12.  3(2a+46)+3(3a-66).    ;  Ans.  3<5a—  26). 

13.  16a*(a+6)+16a2(c-d).  -  Ans.  16a*(a+6+c-d). 

14.  (a-6)»+2a(a-6)t.  Ans.  (a-6)*(3a  -6). 

~jn 

228.  Squares  and  square  roots  of  monomials. — By  the 
principles  of  multiplication  already  given,  the  square  of  a 
monomial  may  be  found  as  follows: 

(1)  The  sign  is  always  plus. 

(2)  The  numerical  coefficient  is  the  square  of  the  numerical 
coefficient  of  the  monomial. 

(3)  The  exponent  of  any  letter  is  twice  the  exponent  of  the 
same  letter  in  the  monomial. 

Thus,  (5o*6s)»  =  25o46«,  ( -4a86*d)1  =  16a«64d*,  and  (iz»i/)J  -  Jz«y*. 

The  square  root  of  a  monomial  can  be  found  by  doing  the 
inverse  processes  to  those  for  finding  the  square  of  a  monomial. 

(1)  The  square  root  can  be  found  of  a  positive  number  only. 

(2)  The  numerical  coefficient  is  the  square  root  of  the  numerical 
coefficient  of  the  monomial. 

(3)  The  exponent  of  any  letter  is  one-half  the  exponent  of  the 
same  letter  in  the  monomial. 

It  follows  that  the  monomial  of  which  the  square  root  is 
to  be  taken  must  have  a  numerical  coefficient  that  is  a  perfect 
square  and  all  the  exponents  must  be  even  numbers.  Other- 
wise the  square  root  cannot  be  found  exactly. 

Thus,  Vl6a46»  «=4a*6  and  \/225x*y*z*  -  15zVz;  but  VlOa46«  can  only 
be  expressed  as  \/10al6a;  and  \/35o»6  cannot  be  found. 


DIVISION,  SPECIAL  PRODUCTS,  AND  FACTORS      279 

EXERCISES  71 

Find  the  indicated  square  or  square  root  of  the  following  when  possible. 

When  not  possible,  tell  what  change  in  the  expression  would  make  it 
possible. 

1.  (3ab3)2.  8.  (-20x6?/4)2.             15.  -y/576*4?/2. 

2.  (-2x2y2)2.  9.  Vl6»y.                 16.  \/-16*6z4. 

3.  (-9ac2d)2.  10.  -y/25m4n6.                 17.  \/x*y*z*. 

4.  (£am3r)2.  11.  \/13a624.                   18.  \/3s2r/6. 
6.  (-icVz)2-  12.  V256XV0-               19.  yVfl2r2. 


6.   (llm'n'z)8.  13.  Vol^T2.  20 


.   \(f  ) 


7.   (z3?/425)2.  14.  v/49ac64z6.  21.  -V/f-^-j  Vg2. 

229.  The  square  of  a  binomial. — By  multiplication 

(a+b)2  =  a2+2ab+b2, 

(a-b)2  =  a2-2ab+b2. 

Here  a  and  b  are  general  numbers,  so  we  may  use  the  state- 
ments as  formulas  to  find  the  square  of  the  sum  or  the  differ- 
ence of  any  two  numbers.  These  formulas  may  be  translated 
into  words  as  follows: 

(1)  The  square  of  the  sum  of  two  numbers  equals  the  square 
of  the  first  plus  twice  the  product  of  the  first  by  the  second  plus 
the  square  of  the  second. 

(2)  The  square  of  the  difference  of  two  numbers  equals  the 
square  of  the  first  minus  twice  the  product  of  the  first  by  the  second 
plus  the  square  of  the  second. 

The  use  of  these  principles  will  save  much  work  in  multi- 
plication. 

Example  1.     Find  the  value  of  (cd+e)2. 

The  square  of  the  first  term  =  (cd)2  =  c2d2. 

Twice  the  product  of  the  first  by  the  second  =  2(cd)e  =  2cde. 

The  square  of  the  second  term  =  e2. 


Example  2.     (2a+62)2  =  4a2+4a&2+64. 
Example  3.     (2z2-37/3)2  =  4z4 


EXERCISES  72 

Write  the  products  of  the  following  without  actual  multiplication,  and 
then  test  by  actual  multiplication. 

1.  (w+2)2.  2.  (a+26)2.    3.  (x+2?/)2.    4.  (3x+y)2. 

5.  (x2+4)2.  6.  (2z2-3)2.  7.  (2x-7)2.    8.  (aa;2-4i/).2 


280  PRACTICAL  MATHEMATICS 


9.  (3oz-4y)«.  Arm. 

10.  (2aV  -3y)«.  Arw.  4aV  -  12a  V 

230.  Factors  of  a  trinomial  square.  —  A  trinomial  square  is  a 
trinomial  that  is  the  square  of  a  binomial. 

Thus,  a*+2a6+6*  is  a  trinomial  square  because  it  is  the  square  of  the 
binomial  a+b.  Its  factors  then  are  evidently  (a+b)(a+b).  Likewise, 
the  factors  of  a2  -  2ab  +61  are  (a  -  6)  (a  -  6)  .  The  factors  of  4z2  -  12z  +9 
are  (2z-3)  (2z-3). 

It  should  be  carefully  noticed  that  a  trinomial  square  has 
two  positive  terms,  each  of  which  is  the  square  of  a  monomial  ; 
and  one  term,  either  positive  or  negative,  that  is  twice  the 
product  of  the  square  roots  of  the  other  two  terms.  If  this 
term  is  positive  the  factors  are  sums,  and  if  negative  the  factors 
are  differences. 

Thus,  9a4—  24a2t/2  +  16y4  is  a  trinomial  square,  for  9a4  and  16y4  are 
each  positive  and  squares  of  the  monomials  3o2  and  4t/2  respectively; 
and  24a2t/2  is  twice  the  product  of  these  square  roots.  The  factors  are 
(3a2-4t/2)(3a*-4i/2). 

Since  by  definition  the  square  root  of  a  number  is  one  of  its 
two  equal  factors,  the  square  root  of  a  trinomial  square  is 
one  of  its  two  equal  factors. 

EXERCISES  73 

Determine  which  of  the  following  are  trinomial  squares.  Factor  and 
find  the  square  root  when  possible. 

1.  x*+2xy+y*.  10.  2562  +  16z2-506z. 

2.  4z2-4z?/+?/2.  11.  4z4+z/4+4z2r/2. 

3.  9z4  +  18zV+9y2.  12.  225a2-270a+81. 

4.  IGz'-Sz^+r/*.  13.  \x*  +  \xy  +  \y*. 

6.  z2-2z-l.  14.  81a«-108aV+36y4. 

6.  z«+4z2+4.  15. 

7.  z»+3z+9.  16. 

8.  25zlo  +  10zs+l.  17.  100+a2+20a. 

9.  14z2+6z+8.  18.  16z2+25j/2-50zj/. 

231.  The  product  of   the  sum  of   two  numbers    by  the 
difference  of  the  same  two  numbers.  —  By  multiplication 

(a+b)(a-b)=a2-b2. 

Since  a  and  b  are  general  numbers,  we  may  use  this  state- 
ment as  a  formula  and  so  write  at  once,  without  actual  mul- 
tiplication, the  product  of  the  sum  and  the  difference  of  any 


DIVISION,  SPECIAL  PRODUCTS,  AND  FACTORS      281 

two  numbers.  The  formula  may  be  translated  into  words 
as  follows: 

The  product  of  the  sum  and  the  difference  of  two  numbers 
equals  the  difference  of  their  squares. 

Example  1.     (2c+3&)(2c-36)  =4c2-962. 

Example  2.     (16  +2)  (16  -2)  =  162-22  =  256-4  =  252. 

Example  3.    102X98  =  (100+2)  (100  -2)  =  10,000-4  =  9996. 

EXERCISES  74 

Find  the  product  of  the  following  without  actual  multiplication  and 
test  by  actual  multiplication: 

1.  (2+2y)(2-2y).  2.  (3x-y)(3x+y). 

3.  (12z-  13)  (12s  +  13).  4.  (16z2y-2)(16z22/+2). 

5.  (x3+ys)(x3-ys).  6.  (3<2-40(3<2+4<). 

7.  (4zV  +  l)(4z22/3-l)-  8.  (2-33-5)(2-33+5). 

9.  [0  +  (&+l)][a-(6  +  l)].  10.  [ 
11.  (2c+d+e)(2c+d-e). 
Suggestion.     This  may  be  written 


12.  (a+b-2c)(a+6+2c). 

13.  (x2-y*-xy)(x2+ 
Suggestion.     This  may  be  written 

[x2  -(y2+xy')][x2  +  (y2  +xy)]  =x*-(y2+xy*)2  =etc. 

14.  Cc2-y2-z?/)(>2-2/2+zy). 
16.  (7+x+y)(7-x-y). 

16.  (3-x+y)(3+x+y). 
Suggestion.     This  may  be  written 

[(3+y)  -x][(3+^)  +x]  =  (3+?/)2-x2  =etc. 

17.  (o-y-2)(o-y+z).  18.  [4a-(*-2&)][4a+(«-2&)]. 
19.  (Ha;3?/-z4)(llx3?/+24).                 20.   (30x2?/3+3a;2/)(30x2?/3-3x?/). 
21.  95X105.                     22.  995X1005.  23.  64+56. 

24.  75X85.  26.  505X495.  26.  706X694. 

232.  Factors  of  the  difference  of  two  squares.^  —  From  a 
consideration  of  the  preceding  it  is  easily  seen  that  the  difference 
of  two  squares  can  be  factored  into  two  binomial  factors  that 
are,  respectively,  the  sum  and  the  difference  of  the  square 
roots  of  these  squares. 

Example  I.    4-a2=(2+a)(2-a). 

Example  2.     16a4-9?/=  (4a2+3t/)(4a2-3?/). 

Example  3.     (a+6)2-22=(a+6  +  2)(a  +  fe-2). 

Example  4.  a2-62+26c-c2  =a2-(62-26c+c2)  =a2  -(&- 
c)2=(a+6-c)(a-&+c). 


282  PRACTICAL  MATHEMATICS 

EXERCISES  75 

Factor  the  following  and  test  by  multiplication : 

1.  4-x».  2.  16-4y».                     3.  o»-l. 

4.  l-9x«.  5.  Sla'-lGfc*.                6.  7* -5*. 

7.  225-64a».  8.  36a'-496*.                9.  4a16» 

10.  64a«&«-l(KW*.  11.  3>a<-2»&«.               12.  2*  34<f*-5V. 

13.  (x+y)»-«».  Ana.  (x+y+z)(x+y-z). 

14.  (a+6)'-64.  4rw.  (a+6+8)(a+6-8). 

15.  x*+y*-2xy-z*.  Ana.  (x-y+z)(x-y-z). 

16.  25-(a+6)».  Ana.  (5+a+6)(5-a-6). 

17.  25-(a-6)».  Ana.  (5+a-6)(5-a+6). 

18.  P*-4PC+4C1-C*.  Ana.  (P-C)(P-3C). 

19.  (3a-6)»-(2a+26)».  4rw.  (5a+6)(a-36). 

20.  (2x-7)'-(3x-l)J.  4n*.  (5x-8)(-x-6). 

233.  The  product  of  two  binomials  having  one  common 
term. — By  multiplication,  we  find  the  following  products: 

(1)  (a+2)(a+3)=a2+6a+6. 

(2)  (a-2)(a-3)  =  a2-6a+6. 

(3)  (a+2)(a-3)=a2-a  -6. 
.    (4)  (a-2)(a+3)  =  a2+a  -6. 

(6)  (a+b)(a+c)=a2  +  (b+c)a+bc. 

From  an  inspection  of  these,  the  truth  of  the  following 
statement  can  be  seen  : 

The  product  of  two  binomials,  having  one  common  term  and 
the  other  terms  unlike,  is  a  trinomial  consisting  of  the  square 
of  the  common  term,  the  algebraic  sum  of  the  unlike  terms  times 
the  common  term,  and  the  product  of  the  unlike  terms. 

Thus,  in  (1)  above,  the  common  term  is  a  and  the  unlike  terms  are  2 
and  3.  The  square  of  the  common  term  is  a1.  The  algebraic  sum  of 
the  unlike  terms  is  2+3  =5,  and  this  times  the  common  term  is  5  times 
a  =  5a.  The  product  of  the  unlike  terms  is  2X3=6.  Hence  the  result 
(a+2)(a+3)=a'+oa+6. 

Likewise  in  (3),  the  square  of  the  common  term  is  a*.  The  algebraic 
sum  of  the  unlike  terms  is  the  sum  of  +  2  and  —3,  or  —  1.  This  times  the 
common  term,  a,  is  — o.  The  product  of  the  unlike  terms  is  2  X  —3  =  — 6. 
Hence  the  result  (a+2)(a-3)  =at-a-6. 

EXERCISES  76 

Find  the  products  of  the  following  without  actual  multiplication  and 
test  by  actual  multiplication: 

1.  (x+3)(z+4).  2.  (z-4)(x+3).  3.  (x-3)(x+4). 

4.  (x-3)(x-4).  5.  ^x+7)(x+8).  6.  (x-7)(x-8). 

7.  (;r+7)(x-8).  8.  (x-7)(x-f-8).  9.  (2n-4)(2n+5). 


DIVISION,  SPECIAL  PRODUCTS,  AND  FACTORS      283 

10.  (2n+4)(2n+5).  11.  (x+l)(x-ll). 

12.  (r- 16)(r +17).  13.  (z2+6)(z2-5). 

14.  (2z+7)(2z-18).  16.  (6o+3)(5o-6). 

16.  (6+8) (6 -5).  17.  (xy+2)(xy-3). 

18.  (3+a^)(4+xy).  19.  (2z+3)(4z+3). 
20.  (3z-4)(3a;+9). 

234.  To  factor  a  trinomial  into  two  binomials  with  one  com- 
mon term. — By  a  careful- study  of  the  preceding  exercises,  we 
may  determine  how  to  proceed  in  factoring  such  trinomials  as 
the   products  in  those  exercises.     The  method  of  factoring 
those  forms  will  best  be  seen  by  considering  examples. 

Example  1.     Factor  «2+9a+20. 

This  has  one  term,  a2,  that  is  a  perfect  square;  a,  the  square 
root  of  this,  is  to  be  the  common  term  of  the  factors,  if  there 
are  any.  The  unlike  terms  of  the  factors  must  have  a  product 
of  +20  and  a  sum  of  +9.  By  inspection  we  see  that  +5 
and  +4  have  such  a  product  and  sum.  Hence  the  factors 
of  o2+9a+20  are  (a+5)(o+4). 

Example  2.     Factor  a2  — a— 20. 

As  before,  the  common  term  is  a.  The  unlike  terms  have 
a  product  of  —20  and  a  sum  of  —1.  The  product  being  — , 
one  of  the  terms  is  —  and  one  + .  The  sum  being  — ,  shows 
that  the  larger  in  absolute  value  is  — .  This  gives  —5  and 
+4  as  the  numbers. 

Hence  a2- a -20  =  (a- 5) (a+4). 

235.  Other  forms. — Many  trinomials  that  appear  to  be  of 
the  kind  here  considered  cannot  be  factored  in  this  way. 

For  instance,  z2+7z+5  cannot  be  factored  as  here,  for  we  can  find  no 
integral  numbers  which  have  a  sum  of  7  and  a  product  of  5. 

There  is  still  another  class  of  trinomials,  those  where  the 
binomial  factors  have  no  common  term,  and  many  other 
expressions  which  can  be  factored.  These  will  not  be  taken 
up  here,  either  because  they  are  too  difficult  or  because  they 
are  of  less  use  than  the  ones  considered. 

EXERCISES  77 

Factor  the  following  if  possible ;  if  not,  change  a  term  so  that  they  may 
be  factored.  Test  your  work  by  multiplication. 

1.62-76  +  12.  2.  62  +  116+30.  3.  c2-c-30. 

4.  z2+2z-8.  5.  a2-3z-10.  6. 


284 


PRACTICAL  MATHEMATICS 


7.  ?/'-W- 


8.  a»+3a  +  14. 


26. 


10.  r2+20r+64. 

13.  a»+9a-10. 

16.  x*+6a-72. 

19.  x»-12x-45. 

22.  nl  +  15n-16. 

26.  a»-lla-60. 

28. 

80. 

32. 

34.  aJ+3a6-546J. 

36.  z'-(a+6)x+a&. 

38.  (m-n)'-f7(m-n)+12. 

Factor  the  following  expressions  by  the  methods  given  : 


14.  a»-a-132. 
17.  x2-x-90. 
90.  a»+4a-21. 
23.  n'  +  13n+36. 
az-|-10a-24. 
29. 


9.  x»-15z+56. 

12.  x»-3x  +  16. 
45.  a1  -a  -72. 
'  18.  x*  +  15z-34. 

21.  a»-7o-33. 

24.  n*-6n-60. 


33.  as-14a+40. 
36.  l-lSy-SSy 
37. 
39. 


40.  xV-16. 
43.  aV-&V- 
46.  100z4-y«. 
49.  64x9-16z7. 
62.  (x 
64.  al 
66.  x2 


41.  xt/'-4x. 


47. 
60. 


42.  x 

46.  3ax»-18ax-180a. 
48.  9x*+4j/*  +  12xy. 
61.  64x*  +  16x7+x«. 


16a4-16x4. 
63. 
66. 

67.  (x-t/)2-2ac(x-y)+a2c2. 
The  following  are  some  of  the  answers  of  the  above  exercises.     They 
should  be  consulted  only  after  the  factors  are  found. 

4.  (x+4)(x-2).       12.  No  factors.  16.  (a  +8)  (a-  9). 

28.  (ab-15)(a6-l).31.  (x+3r/)(x+9y).  32.  (l+7x)(l-2x). 

36.  (x-a)(x-6).       38.  (m-n+4)(m-n+3)44.  fcx(z+a)(x-a). 
46.  3a(x2-6z-60).  63.  (&-c+8)(6-c+8).    66. 


CHAPTER  XXVI 
EQUATIONS 

236.  If  an  equation  has  indicated  multiplications  and 
signs  of  grouping,  it  is  usually  best  to  perform  the  multipli- 
cations and  remove  the  signs  of  grouping  before  proceeding 
with  the  solution  of  the  equation. 

Example  1.     Find  the  value  of  c  from 

4c+3[2c-4(c-2)]  =  72-6c. 

Solution, 

(1)  Given  equation,     4c+3[2c-4(c-2)]=72-6c. 

(2)  Simplifying,  4c+3[2c-4c+8]     =72-6c. 

(3)  Simplifying,  4c+6c-12c+24     =  72-6c. 

(4)  Transposing,          4c+6c-12c+6c     =72-24. 

(5)  Collecting  terms,  4c  =  48. 

(6)  Dividing  by  the  coefficient  of  c,         c  =  12. 
Test,  48+3[24-4(12-2)]  =  72-72,  or  0  =  0. 
Example  2.     Solve  for  x  : 


Solution. 

(1)  Given  equation,  (l+3z)2  =  (5-z)2+4(l-z)(3-2z) 

(2)  Removing  parentheses, 


(3)  Transposing,  9z2-z2-8z2+6z+10:r+20z  = 

(4)  Collecting  terms,  36z  =  36. 

(5)  Dividing  by  36,  re  =  1. 

Test,  (l+3)2=(5-l)2+4(l-l)(3-2), 
or42  =  42+0,  or  16  =  16. 


EXERCISES  78 

1.  7x-5=x-23.  Ans.  -3. 

2.  5x-12  =  6x-8.  Ans.  -4. 
J..  7x  +  19  =  5x+7.  Ans.  -6, 

4.  2x-(5x+5)=7.  Ans.  -4. 

6.  3(x  +  l)  =  -5(x-l).  Ans.  I 

285 


286  PRACTICAL  MATHEMATICS 

$.  7(z-18)=3(z-14).  Ana.  21. 

J.  2(z-l)-3(z-2)+4(z-3)+2-0.  Ans.  2. 

8.   14z  +20  -  12  =  -  20z  +35*.  Ant.  8. 

0y2(z-l)+3(z-2)+4(z-3)-0.  ATM.  $} 

10.  5(2z  +  l)-7=3(2z-7)+51.  Ann.  *. 

11.  3(z+4)(z-2)-5=3(z+5)(z-3)+z.  ATM.  16. 

12.  (z+2)I-z»=z-5.  Ana.   -3. 

13.  (z-4)(z+4)=(z-6)(z+5)+25.  Arw.  fa 
It.  3(z  +  l)-2(2z+5)=6(3-z).  Ana.  5. 

15.  lla  =  3(z-2a)-5(2z-2o).  Ans.   -a. 

16.  3(26-4z)-(z-6)--66.  Ans.  b. 

17.  5(4z-3a)-6(3z-2a)=3a.  Ans.  3a. 
Find  values  of  a  in  exercises  18  to  24. 

18.  3  +  (a+4)2  =  (a+3)2-4a  +  17.  Ans.  I 

19.  2.5a-6.75  =  1.25a-3.  Ans.  3. 

20.  8a-12=6a+4.  Ann.  §.-" 
21»  37a-(4+7)=41a+25.  Xn*.   -9X 
22.  12.75a  +6.25  =7.25a  +  17.25.  ^Irw.  2. 

J3-  7(25-o)-2a=2(3a-25).  An«.  16^^ 

24.  5a-17+3a-5=6a-7-8a  +  115.  Ans.  13. 
Find  values  of  y  in  exercises  25  to  31. 

25.  2(t/-l)-3(i/-2)=4(3-!/)-2.  Ans.  2.' 

26.  5r/-6(y+l)-7(j/+2)-8(i/+3)=0.  Ans.   -2J. 


28.  (7/  +  l)«-(t/»-l)=y(2j/  +  l)-2(r/+2)(i/  +  l)+20.  Ans.  2. 

29.  6(r/*-3y+2)-2(j/z-l)=4(y+l)(7/-f2)-24.  Ans.  1. 

30.  2y-5[3j/-7(4j/-9)]=66.  ATW.  3. 

31.  3(5-6z/)-5(i/-fi[l-3t/  +  15])=23.  Ans.  4. 
Find  values  of  z  in  exercises  32  and  33. 

32.  84  +  (z+4)(z-3)(z+5)  =  (z  +  l)(z+2)(z+3).  Ans.  1. 

33.  (z  +  l)(z+2)(z+6)=z'+9zI+4(7z-l).  Ans.  2. 

34.  Find  two  numbers  whose  difference  is  25  and  whose  sum  is  4| 
times  their  difference.  Ans.  40  and  65. 

35.  A  rectangular  field  is  5  rods  longer  than  it  is  wide.     If  it  was 
2  rods  wider  and  3  rods  shorter  it  would  contain  4  square  rods  less.     Find 
dimensions  of  the  rectangle.  Ans.   13  rd.  long,  8  rd.  wide. 

Solution.     Let  z  =  number  of  rods  in  width. 
Then  z  +5  =  number  of  rods  in  length, 

and         z(z+5)  =  number  of  square  rods  in  field. 
Also  z+2  =number  of  rods  in  width  of  second  field, 

and  z+2  =  number  of  rods  in  length  of  second  field. 

Then  (z+2)  (z+2)  =  number  of  square  rods  in  second  field 
/.z(z+5)-(z+2)(z+2)=4. 
z*+5z—  z1  —  4z  —  4  "4. 
j*-z»+5z-4z-4+4. 

z  =  8. 
z+5-13. 


EQUATIONS 


287 


36.  The  difference  between  the  squares  of  two  consecutive  numbers  is 
25.     What  are  the  numbers?  Ans.   12  and  13. 

37.  The   difference   between   the   squares   of   two  consecutive  even 
numbers  is  84.     What  are  the  numbers?  Ans.  20  and  22. 

38.  The  height  of  a  flagstaff  is  unknown;  but  it  is  noticed  that  the  flag 
rope,  which  is  4  ft.  longer  than  the  staff,  when  stretched  out  just  reaches 
the  ground  at  a  point  25  ft.  from  the  foot  of  the  staff.     If  the  ground  is 
level,  find  the  height  of  the  staff.  Ans.  76  J  ft. 

Suggestion.     Fig.  198  shows  the  rope  stretched  to  a  point  25  ft.  from 
the  foot  of  the  staff.     This  makes  a  right  triangle  which  has  the  rope 
as  hypotenuse.     We  then  have  the  equation 
(z+4)2=252+.r2. 


4- 


i<  -25  ffc— >t 
FIG.   198. 


A\<— 40-ft— 
FIG.   199. 


39.  A  flagstaff,  CD,  Fig.  199,  75  ft.  high,  breaks  at  point  B  and  end 
D  strikes  at  A,  a  distance  of  40  ft.  from  C.     Find  the  length  BD  that 
was  broken  off.  Ans.  48?  ft. 

Suggestion.     Let  BD=x,  then  CB  =  75—x. 

40.  A  piece  of  sheet  iron  containing  625  sq.  in.  is  bent  into  a  cylinder 
9  in.  in  diameter.     How  high  is  the  cylinder?     What  is  its  volume? 

Ans.  22.1  in.;  1406+  in.3 

41.  At  what  rate  simple  interest  will  $75.00  amount  to  $106.50  in 
6  years?  Ans.  7%. 

237.  Equations  solved  by  aid  of  factoring. — The  equations 
considered  so  far  have  reduced  to  a  form  in  which  a  certain 
number  of  times  the  unknown  equaled  some  number.  Thus, 
6z  =  12  is  such  a  form.  They  are  called  simple  equations. 

All  equations  do  not  reduce  to  such  a  form  as  this.  For 
instance,  when  the  equation  has  been  reduced,  we  may  have 
an  equation  in  which  the  square  of  the  unknown  equals  some 
number.  Thus,  £2  =  5  is  such  a  form.  Such  an  equation  is 
called  a  pure  quadratic  equation. 

Again,  when  the  equation  is  simplified  and  reduced,  we 
may  have  a  form  containing  the  square  and  the  first  power 


Jvs  PRACTICAL  MATHEMATICS 

of  the  unknown  equaling  some  number.  Thus,  x1— 5x  =  24 
is  such  a  form.  Such  an  equation  is  called  an  affected  quad- 
ratic equation. 

Some  of  these  forms  of  equations,  together  with  certain 
other  forms,  can  be  solved  by  the  aid  of  factoring. 

Example  1.     Solve  the  equation  x2— 5x+6  =  0. 

Discussion.  This  equation  puts  the  question:  For  what 
values  of  x  does  x2— 5x+6  equal  zero?  If  we  factor  the 
expression  in  the  first  member  we  get  (x— 2)(x— 3)  =0. 
The  question  now  is:  For  what  values  of  x  does  the  product 
(x-2)(x— 3)  have  the  value  zero?  We  know  that  the 
product  of  two  factors  is  zero  if  either,  or  both,  factors  are  zero 
and  not  otherwise.  Hence  the  product  is  zero,  if  x  —  2  =  0,  or 
if  x  —  3  =  0.  Thus,  the  solution  of  x2— 5x+6  =  0  depends  upon 
the  solution  of  the  two  simple  equations,  x  —  2  =  0  and  x  —  3  =  0. 
These  give  the  values  2  and  3  for  x. 

That  these  are  the  values  of  x  may  be  tested  by  substituting 
each  one  separately  in  the  equation 
x2-5x+6  =  0. 

Substituting  x  =  2,  gives  4-10+6  =  0,  or  0  =  0. 

Substituting  x  =  3,  gives  9-15+6  =  0,  or  0  =  0. 

The  values  of  the  unknown  number  that  satisfy  the  equation, 
that  is,  answer  the  question,  are  called  roots  of  the  equation. 

A  quadratic  equation  having  one  unknown  letter  always  has 
two  roots. 

Example  1.     Solve  the  equation  x2— 25  =  0. 

First  solution. 

(1)  Given  equation,  x2— 25  =  0. 

(2)  Factoring,  (x+5)(x-5)  =0. 

(3)  Putting  each  factor  equal  to  zero, 

x+5  =  0  and  x  — 5  =  0. 

(4)  Transposing,  x  =  —  5  and  x  =  5. 
Testforx=-5,  25-25  =  0. 

Test  for  x  =  5,      25-25  =  0. 
Second  solution. 

(1)  Given  equation,  x2  — 25  =  0. 

(2)  Transposing,  x2  =  25. 

(3)  Taking  the  square  root  of  each  member  of  the  equation, 

x=±5. 


EQUATIONS  289 

Here  the  sign±  is  read  "plus  or  minus."  It  means  that  5 
is  a  plus  as  well  as  a  minus  quantity.  It  should  be  noted  here 
that  we  are  saying  that  25  has  the  two  square  roots,  +5  and 
—  5.  Either  of  these  is  the  square  root  of  25,  for  (+5)2  =  25 
and  also  (  — 5)2  =  25.  Hence  both  fulfill  the  definition  of  a 
square  root,  that  is,  one  of  the  two  equal  factors  into  which  a 
number  may  be  divided. 

Any  positive  number  has  two  square  roots,  one  positive  and 
one  negative,  both  equal  in  absolute  value. 

Example  3.     Solve  (3  +  !)  (3 -3)  (23 -16)  -0. 
Equating  each  factor  to  zero, 

&  +  l=0,  3-3  =  0,  and  23- 16  =  0. 
Solving  these,  x=  —  1,  x  =  3,  and  re  =  8. 

We  have  the  following  rule  of  procedure  when  solving  an 
equation  by  the  aid  of  factoring: 

RULE.     (1)     Simplify   the   equation   as   much   as   possible. 

(2)  Transpose  all  terms  to  the  first  member  of  the  equation. 

(3)  Factor  the  expression  in  the  first  member. 

(4)  Equate  each  factor  to  zero. 

(5)  Solve  each  of  these  equations. 

EXERCISES  79 

Solve  the  following  by  the  aid  of  factoring: 
1.  (x-4)(x-3)=0.  2.  (x-5)(x+6)=0. 

3.  (x-2)(x  +  l)(z+3)=0.  4.  (x  +  l)(x-5)(x-3)=0. 

6.  (2x-l)(x+4)=0.  6.  x(x-2)(3x+5)=0. 

7.  x2-16=0.  8.  (z2-9)(x2-36)=0. 

9.  x2-16=48.  Am.  Sand  -8. 

10.  x2-z  =  56.  Ans.  Sand  -7. 

11.  (2x  +  l)(x+3)=z2-9.  Ans.    -3  and  -4. 
Suggestion.     It  is  necessary  first  to  clear  of  parentheses,  transpose,  and 

unite  before  factoring. 

Clearing  of  parentheses,    2x2+7x+3  =  z2  —  9. 
Transposing,  2x2-x2+7x+3+9  =0. 

Collecting  terms,  x2 +7x  + 12  =  0. 

Factoring,  (x +3)  (x  +4)  =  0. 

12.  If  24  is  added  to  the  square  of  a  number,  the  sum  equals  10  times 
the  number.     Find  the  number.  Ans.  4  or  6. 

13.  If  78  be  subtracted  from  the  square  of  a  number,  the  difference 
equals  7  times  the  number.     Find  the  number.  Ans.   13  or  —6. 

14.  If  3  is  added  to  a  number,  the  square  of  the  sum  is  9  more  than  13 
times  the  number.     What  is  the  number?  Ans.  0  or  7. 

19 


290  PRACTICAL  MATHEMATICS 

16.  A  rectangle  is  8  in.  longer  than  it  is  wide.  Find  the  dimensions, 
if  the  area  is  240  sq.  in.  Ant.  12  in.  by  20  in. 

16.  Find  the  dimensions  of  a  rectangle  that  is  4  rods  longer  than  it  in 
wide  if,  when  the  length  is  increased  by  6  rods  and  the  width  by  4  rods, 
the  area  is  doubled.  Ana.  10  rd.  by  14  rd. 

17.  The  base  of  a  triangle  is  3  in.  longer  than  the  altitude,  and  the 
area  is  44  sq.  in.     Find  the  length  of  the  base  and  the  altitude. 

Ant.  11  in.  and  8  in. 

18.  The  altitude  of  a  triangle  is  3  times  the  base,  and  the  area  is 
37  J  sq.  in.     Find  the  length  of  the  base  and  the  altitude. 

An*.  5  in.  and  15  in. 

238.  Formulas. — A  formula  as  given  usually  stands  solved 
for  one  letter  in  terms  of  several  others.  For  instance,  formula 
[34]  is  T  =  ph+2A.  Here  T  is  stated  in  terms  of  p,  h,  and  A. 

It  often  happens  that  one  wishes  to  express,  say,  h  in  terms 
of  T,  p,  and  A.  To  do  this  it  is  only  necessary  to  solve  the 
formula  as  an  equation,  and  find  the  value  of  the  particular 
letter  desired  in  terms  of  the  others. 

Example.  Solve  the  formula  T  =  ph+2A  for  each  of  the 
other  letters. 

Solution.  Here  there  are  three  other  letters  than  T,  and 
we  will  solve  for  p,  h,  and  A  in  turn. 

(1)  Given  equation,  T  =  ph+2A. 

(2)  Transposing,  -p/i=-T+2A. 

(3)  Dividing  by  the  coefficient  of  p  which  is    —h,    and 
indicating  the  division,  since  it  cannot  be  performed, 

T-2A 
p  —  — 7 Ans. 

(4)  Solving  (2)  for  h,  since  it  is  properly  transposed, 

,      T-2A     . 

h  = Ans. 

P 

(5)  To  solve  for  A  transpose  (1),  -2A  =  —  T+ph. 

rrt L 

(6)  Dividing  by  the  coefficient  of  A,  A  = — « — ••  Ans. 

EXERCISES  80 

In  the  following,  the  numbers  in  the  brackets  are  the  numbers  of  the 
formulas  as  given  in  previous  chapters,  where  their  meaning  can  be  found. 
The  ability  to  do  such  problems  as  these  is  very  important. 

1.  [32]  A  -rob.     Solve  for  a  and  6.  Am.  ~;  ~ 

2.  [33]  S  =  ph,     Solve  for  p  and  A.  Ant.  |;  — 


EQUATIONS  291 

e 

3.  [42]  S  =  2irrh.     Solve  for  r.  Ans.  ^-^ 

V 

4.  [43]  F  =  7rr2/t.     Solve  for  h.  Ans.  —  2- 

v 

6.  [46]  V  =  irRzh—nr*h.     Solve  for  h.  Ans.      /p2_  2)- 

6.  [69]  A=47r2flr.     Solve  for  R.  Ans.  —  - 

47r2r 

^ 

7.  [67]  Z  =  2irrh.     Solve  for  r.  Ans.  -^-^ 

8.  Using  the  answer  of  exercise  7,  find  the  radius  of  a  sphere  on  which 
a  zone  of  altitude  3  ft.  has  an  area  of  32  sq.  ft.         Ans.   1.698—  ft. 

T        IT 

9.  [35]  71  =  6a2.     Solve  for  a2  and  then  for  a.  Ans.  -',   x/- 


10.  Using  the  second  answer  of  exercise  9,  find  the  edge  of  a  cube 
whose  total  surface  area  is  3258  ft.2  Ans.  23.302  -f  ft. 


11.  [60]  7  =  2ir2#r2.     Solve  for  r.  Ans. 

12.  [55]  S  =  47rr2.     Solve  for  r.  Ans.    \- 

13.  Using  the  answer  of  exercise  12,  find  the  radius  of  a  sphere  that 
has  a  surface  area  of  2756  ft.2  Ans.   14.82- ft. 

/F 

14.  [43]  7=7rr2/i.     Solve  for  r.  Ans.    A/^- 

\  TTrt 

15.  Using  the  answer  of  exercise  14,  find  the  radius  of  a  right  circular 
cylinder  whose  altitude  is  16  in.  and  volume  2674  in.3     Ans.  7.294  —  in. 

16.  Disregarding  the  resistance  of  the  air,  v  =  \/2gh  is  a  formula  that 
gives  the  velocity  in  feet  per  second  a  body  will  have  after  falling  from  a 
height  of  h  feet.     Solve  this  for  h  and  get  a  formula  for  the  height  to 
which  a  body  will  rise  if  thrown  upward  with  ««,  velocity  of  v  feet  per 

v2 
second.  Ans.  ft=«— 

r 

Suggestion.     First  square  both  members  of  the  equation,  which  gives 

v*=2gh. 

17.  Use  the  formulas  of  exercise  16  and  find  (1)  the  velocity  a  stone 
will  have  after  falling  125  ft.,  (2)  the  height  to  which  a  stone  will  go  if 
thrown  upward  with  a  velocity  of  200  ft.  per  second,     (g  =  32. 2.) 

Ans.  v  =  89.72+  ft.  per  second;  h  =  62 1.1  ft. 

18.  The  formula  vt  =  vo+32.2t  gives  the  velocity  that  a  falling  body 
will  have. 

t  =  time  in  seconds  the  body  has  been  falling, 

z;o=  velocity  in  feet  per  second  the  body  has  at  the  start,  that  is, 

VQ  is  the  initial  velocity, 
vt—  velocity  in  feet  per  second  after  t  seconds. 

Solve  for  v0  and  for  t.  Ans.  t=  ' 


292  PRACTICAL  MATHEMATICS 

19.  Using  the  formulas  of  exercise  18,  find  (1)  the  time  for  a  falling 
body  to  have  a  velocity  of  600  ft.  per  second  if  it  started  with  a  velocity 
of  40  ft.  per  second;  (2)  the  initial  velocity  in  order  that  a  fulling  body 
may  have  a  velocity  of  340  ft.  per  second  after  falling  5  seconds. 

Ana.  17.39+  seconds;  179  ft.  per  second. 

20.  Use  the  formula  given  in  exercise  5  and  find  the  height  of  a  hollow 
cylinder  having  outer  and  inner  radii  of  14  in.  and  10  in.  respectively, 
that  it  may  have  a  volume  of  4.211  cu.  ft.  Ana.  24.13—  in. 

21.  Given  the  formula  V  =  far2h=  1.0472  r*h,  for  finding  the  volume  of 

V  I      V 

a  circular  cone,  solve  for  h  and  for  r.  Ans.  L0472r,;    \fo472li' 

22.  Using  the  formulas  of  exercise  21,  find  (1)  the  altitude  of  a  circular 
cone  having  a  volume  of  800  cu.  in.  and  a  radius  of  8  in. ;  (2)  the  radius  of 
the  base  of  a  circular  cone  having  a  volume  of  456  cu.  in.  and  an  altitude 
of  10  in.  Ans.  11.94-  in.;  6.6- in. 

23.  In  the  formula  given  in  the  answer  to  exercise  35,  page  240, 
A  =td+b(s-\-n).     Solve  this  for  t,  d,  b,  8,  and  n  successively. 

Aru.   t  =  A-b(s+n),  s  =  A-td-bn 
d  b 

24.  Using  the  formulas  of  exercise  23,  find  (1)  t  when  A  =3.35  in.1, 
6=2.04  in.,  s=0.22  in.,  n=0.56  in.,  and  d  =  8  in.;  (2)  6  when  d  =  10  in., 
t  =0.24  in.,  n  =0.63  in.,  s  =0.24  in.,  and  A  =4.45  in.1;  (3)  s  when  d  =  5  in., 
<=0.19in.,  6  =  1. 56  in.,  n  =  0.45in.,  and  A  =1.95  in.2 

Ans.  0.22  in. ;  2.36  in. ;  0. 19  in. 

25.  In  finding  the  area  of  a  trapezoid,  2A  =  (bi+bt)h.     Find  61  if 
A  =  400  in.2,    6j  =  15  in.,  and  h  =  20  in.  Ans.  25  in. 

26.  Find  the  volume  of  a  sphere  that  has  a  surface  of  201.0624  sq.  in., 
the  formula  for  the  volume  of  a  sphere  being  V  =  Jur3.     Use  formula  of 
exercise  12.  Ans.  268.08  cu.  in. 

27.  In    reckoning    simple    interest,    A=prt+p,    where    A  =  amount, 
p  =  principal,  <=time  in  years,  and  r  =  rate  per  cent.     Solve  for  each 

A  A—p          A  —p 

letter.  Ans.  P  =  I-T-T;   *  = i  r  =  — 

1  +rt '  pr  pt 

28.  Using  the  formulas  of  exercise  27,  find  (1)  I  when  p  =$250,  r  =6%, 
and  A  =$300;  (2)  r  when  *=3  years,  p  =  $328,  and  A=  $377.20;  (3) 
p  when  A  =$500,  t=5  years,  and  r  =  4%. 

Ans.  3J  years;  5%;  $416.67. 


CHAPTER  XXVII 
FRACTIONS 

239.  The  same  names  and  terms  are  used  when  referring  to 
fractions  in  algebra  as  are  used  in  fractions  in  arithmetic,  and 
these  terms  have  the  same  meanings.     The  same  principles 
are  applied  and  the  same  operations  performed  as  in  arithmetic. 

Here  only  those  processes  will  be  considered  that  are  neces- 
sary for  the  understanding  of  what  follows. 

When  in  doubt  about  an  operation  in  algebra,  carry  out  a 
similar  operation  using  numerical  numbers,  and  from  this 
determine  what  the  operation  with  the  algebraic  expressions 
should  be. 

REDUCTION  OF  A  FRACTION  TO  ITS  LOWEST 

TERMS 

240.  A  fraction  is  in  its  lowest  terms  when  there  is  no  factor 
common  to  both  numerator  and  denominator. 

Example  1.     Reduce  |~f-£  to  its  lowest  terms. 


Process.     —-  =  —-=. 
\.Zi\j      p'p'o      o 

Here  each  term  of  the  fraction  is  factored  and  then  the  com- 
mon factors  are  cancelled.  We  handle  an  algebraic  fraction 
in  the  same  way. 

A'V  2^1  3 

Example  2.     Reduce        4  4  to  it  lowest  terms. 


6*y 

2 


-x-y^.yy    2x2y 

^  —  y- 

Example  3.     Reduce  ~TT~O~     ~~9  to  its  lowest  terms. 
x2+2xy+y 

Process  X*  ~  ^          &^(x~  I/)     x  ~  V 

Examvle  4 

2 


293 


291 


PRACTICAL  MATHEMATICS 


EXERCISES  81 
Reduce  the  following  fractions  to  their  lowest  terms: 


125 
225 
135 
150' 
28a»x4 


35a»x* 

45a5c»x* 


Ans. 


5mr/2 


141 
2SS' 
21a'x» 
24a»x4' 


Am. 


Ans.  3 — 
Sax 


12a»x» 


Ans. 
Ans. 


3b4 

5cx*y 

1 


9. 
10. 
11. 
12. 
13. 
14. 
16. 
16. 
17. 
18. 
19. 
20. 
21. 
22. 


15a»c»z 


1296xVz» 
792zV 
81zV  ' 


3ax* 
Am.  3a-i. 

Ans.  —• 
6y4 

88 
Ans.       - 


x(z-y)* 
a2-5a+6 
a2-7a+10* 
n*+7n-30 
n2-7n  +  12 

a(x-y)3 


Ans. 


a2+a-12 
25-0* 
a2-!  la  +30* 


a-3 

.  -  =• 
a  —  5 

n  +  10 


x+y 
5a 


.          :    T  • 

a+4 
5+a 


a(x+y)+c(x+y) 
x(a+c)+y(a+c) 


ax8  —  2ax—  8a 
ax2  —  ax—  6a 


,  y~x~z 

" x+y+z 
x-4 


A  na. 


x-3 


REDUCTION  OF  FRACTIONS  TO  COMMON 
DENOMINATORS 

241.  In  arithmetic,  fractions  are  reduced  to  a  least  common 
denominator  before  adding,  so  in  literal  fractions  we  change 
the  fractions  to  fractions  having  the  lowest  common  denomi- 
nator before  adding  them. 


FRACTIONS  295 

The  lowest  common  denominator,  L.  C.  D.,  is  the  lowest 
common  multiple,  L.  C.  M.,  of  the  denominators  of  the  frac- 
tions. We  must  then  first  consider  the  finding  of  the  L.  C.  M. 
of  algebraic  expressions. 

242.  Lowest  common  multiple.  —  Example  1.  Find  the 
L.  C.  M.  of  24,  32,  and  40. 

First  separate  into  prime  factors,  and  then  find  a  number 
which  contains  all  the  factors  of  each. 

Process.     24  =  23-3. 
32  =  25. 
40  =  23-5. 
/.L.  C.  M.  =  25-3-5  =  480. 

Remark.  The  L.  C.  M.  may  also  be  found  by  the  method  of 
Art.  19,  but  that  method  is  not  as  easily  applied  to  algebraic 
expressions. 

Example  2.    Find  the  L.  C.  M.  of  I2x2y,  IQxy3,  and  24x3y. 

Process.     12x2y  =  22-3-x2-y. 


:.L.  C. 

The  L.  C.  M.  is  found  by  taking  each  factor  the  greatest 
number  of  times  it  is  found  in  any  expression. 

Example  3.     Find  the  L.  C.  M.  of  x-+2xy+y-  and  x2-y'2. 
Process. 


/.L.  C.  M.      =(x+yY(x-y}. 

EXERCISES  82 

Find  the  L.  C.  M.  of  the  following: 

1.  72,  288,  64.  Ans.  576. 

2.  576,  256,  128.  Ans.  2304. 

3.  5a3b2,  10a2b3,  25a2fc.  Ans.  50a363. 

4.  4c2,  2ab,  9cd2.  Ans.  36a6c2d2. 
6.  x*-yz,  x2-2xy+y2.  Ans.  (x-y)2(x+y). 

6.  x*-llx+3Q,  z2-12z+35.  Ans.  (z-5)(z-6)(z-7). 

7.  as-a&2,  (a+6)2,  (a-6)2.  Ans.  a(a+6)2(a-6)2. 

8.  a*b+ab2,  a2+2ab+b*.  Ans.  ab(a+b)*. 

9.  x2+7x,  x*+8x+7.  Ans.  x(x  +  l)(x+7). 
10.  o2+3a+2,  a2  -4,  a2-!.  Ans.  a4-5a2+4. 

243.  Fractions  having  a  L.   C.  D.  —  Example  1.     Change 
S,  ^f>  and  H  to  equivalent  fractions  having  a  L.  C.  D. 


290  PRACTICAL  MATHEMATICS 

The  L.  C.  D.  is  found  by  the  method  of  the  preceding  article 
to  be  96.  Now  multiply  both  numerator  and  denominator 
of  each  fraction  by  such  a  number  as  will  make  the  denomina- 
tor in  each  case  96.  How  is  the  multiplier  in  each  case  found  ? 
9  9X654 


SS' 


16  ~  16X6"  96' 


24     24X4     96' 


32     32X3     96' 

Example  2.     Change  ~^  y*+4y-12'  and  y^y  to   frac~ 
tions  having  a  L.  C.  D. 
Process.     By  the  preceding  article  the  L.  C.  D. 


_     x-y(y+6)  _  _ 


y-2     (y- 


vy  —  2v 
~ 


The  multiplier  in  each  case  is  found  by  dividing  the  L.  C.  D. 
by  the  denominator  of  the  fraction.  The  division  is  most 
easily  performed  by  striking  out  those  factors  in  the  L.  C.  D. 
which  are  found  in  the  denominator  of  the  fraction  considered. 
For  this  reason  it  is  best  not  to  multiply  together  the  factors 
of  the  L.  C.  D.  Thus,  in  the  above  the  L.  C.  D.  is  left  in  the 
foim  y(y  — 2)(y+6)  during  the  process,  instead  of  in  the 
foHn  y3+4y2— 12y. 

EXERCISES  83 

Reduce  the  following  to  equivalent  fractions  having  a  L.  C.  D.: 
7      9    41  .        42    J27    JB2 

26'  52'  718'  *'  156'  156'  156' 

JL     4     _5_  A       — °*      8q     _f) 
4o'  6a2'  12a3'  ''  12av  12a3'  " 

_       2         3  2a-26 

3.       rr> r  Ant. 


'  a+b   a-b  "  a'-b1     a*-J 

.       a          /  o*  t__  ai+a1  x*+ax       a* 

x  —  a   x  —  a   x*  —  a* 
3  5 


6. 


a*+3a+2     a*_2a-3 

3o-9  5a  +  10 

Ans. 


(a+l)(o+2)(a-3)    (a-r-l)(a+2)(o-3) 


6. 


FRACTIONS  297 

2x          b  3x2  562 


24x(x+b)     -6b(x+b)      Ox2 -lOb2 

Am"  12(x2-b2)'    12(x2-52)'  12(x2-62)'  12(x2-62)' 


_ 

7.  a+x,  --  .•  Ans. 


a2-x2      a; 

.  —    —  >  -  •• 

a  —  x  a—x     a—x 

0  a2+x2  a2-x2    a2+2ax+x2   a?+xz 

8.  a  —  x.  a+x,  -     —  Ans.  —    —  >  -  —  >  -     —  • 

a+x  a+x  a+x  a+x 


ADDITION  AND  SUBTRACTION  OF  FRACTIONS 

244.  Fractions  can  be  added  or  subtracted  as  in  arithmetic, 
by  first  reducing  them  to  fractions  having  a  common  denomi- 
nator, and  then  adding  or  subtracting  the  numerators.  The 
result  should  then  be  reduced  to  its  lowest  terms. 

oc        a  ct^  ~\~  oc^ 

Example  1.     Find  the  sum  of  -   '—,—  —  ,  and  -     —„• 

a  —  x  a-\-x          az—x* 


Process.    L.  C.  D.  =a2  —  x2. 

x  x(a-\-x) 


a  —  x     (a  —  x)  (a+x)      a2  — 
a  a(a  —  x)      _az  — 


a-\-x     (a-\-x)(a  —  x)     a2  — 


a2  —  z2  a2  —  a;2 

Adding  the  numerators,  the  sum  of  the  fractions  is 


a 

a-\-2x 


7    o      v  4.  i 

Example  2.     From  -r—  -  take  , 

a2  —  ax          a2—  a;2 

Process.    L.  C.  D.  =a(a-\-x}(a—  x)  =a3  —  ax2. 
(a+x)  (a+x)    _a2 


a2  —  ax     (a2  —  ax)  (a+x)        a3  —  ax2 
a+2x  _  (a+2x)a  _a2+2ax 


a2  —  x2     (a2  —  x2)a     a3  —  ax2 
Subtracting  the  numerator  of  the  second  fraction  from  the 

2 
numerator  of  the  first,  the  result  is 


ax 

EXERCISES  84 

1.  to  8.  Add  the  fractions  in  the  preceding  set  of  exercises, 
9.  Add  f,  |,  I,  |.  Ans.  3J 

....      ...aa      a      a    a   a 

10.  Add  ,       ,     ,  Ans.  a. 


PRACTICAL  MATHEMATICS 
a  c-b  c+b  6o+7c-b 


12.  Add       -  and 


o 


in     >  u  in—n       ,  n—T  m—r 

13.  Add  -  and  ---  Ant.  - 

mn  nr  mr 

,  x-2  .  3x+4  230z*- 

14.  Add  lx-\  —  ,=-and  8xH  —  =J  —  -An«. 


=  -.  - 

ox  lor 

4<z+3x  4        5a+2  3x+2a-5al. 

16.  From  -       -  take  —  ^  —  Ans.  - 

Aa  o  ••>" 

3o+12x  3x-3o 

16.  From  3x  take  --  =  --  Ans.  —  =  -- 

o  o 

3a+24  ,      7a6-106  12  -4a 

17.  From  —  r  —  take  -  r=  --  Ans. 


-7         1/nn.c r;  siita.          r 

00  0 

3a-46    2a-6-c  ,  15a-4c  a-46                 81a-46 

18.  Combine  — = 5 1 T^ oi —  '  — 84 

.Sx-2..     0       4x+5  16x+23 

19.  From  2xH = —  take  3x ^ —  Ans.  - 

76  42 

,34                        7  7x-10 

20.  Add s   ^,    and  -: — -- — r-='  Ans.  - 

x  — 2   x— 3            x2  — 5x+6  x*  —  5x+6 


111  1  o&*-a*-a*b-6' 

21.   Combine--^-^^-^^  Ans.        ^^ 


22.  Combine    t_Q    VonH — i_i  f~  -LQO'  ^ns' 

23.  Combine    2  H — f^f  — « — TTT' 

Ans.  ; —. 


24.  Combine  - 


A  x-6. 

*• 


a~b  b-a_         21  -2a«-2a6-46* 

r     I~TI    •  •  *-»  ^  i  r~i  •  *~       r*          ATIS.  ~~.  — 
a2  —  6*     az-|-2a6-|-b1     a  —  6  (a+6)*(a  —  h) 


MULTIPLICATION  AND  DIVISION  OF  FRACTIONS 

245.  Multiplication  of  fractions. — As  in  arithmetic,  the  prod- 
uct of  two  or  more  fractions  is  the  product  of  their  numerators 
divided  by  the  product  of  their  denominators. 

If  we  first  cancel  all  factors  common  to  both  numerator 
and  denominator,  the  result  will  be  in  its  lowest  terms  when 
the  multiplying  is  done. 


FRACTIONS  299 

35        45 
Example  1.     Multiply  ^  by  Q-T- 

Process.     ggX^p^X^^^- 

Here  we  cancel  7  and  9  in  both  numerator  and  denominator. 

27 
Example  2.     Multiply  42  by  ^o- 

Process. 


Example  3.     Multiply       *     y 

J 


.2 
x2  —  2xy+y* 

__     _  ^  __     g>"y> 


Process.       --  __  X  _        __  X 

2  — 


(z+T/Xz+y)     (x-y)(x-y)  x3  x3 

246.  Division  of  fractions. — One  fraction  is  divided  by 
another  by  multiplying  the  reciprocal  of  the  divisor  by  the 
dividend. 

The  reciprocal  of  a  number  is  1  divided  by  that  number. 
The  reciprocal  of  a  fraction  is  then  the  fraction  inverted. 

c       6 
Thus,  the  reciprocal  of  4  is  5,  of  f  is  |,  and  of  r  is  -• 

_.   .,    x-—  llx  —  26,      x2- 
Example.     Divide  -5—5- — ^  by  —„ 


*.+„     26     a;2  —  lSz  +  65 
Process.  =--. — 


x2  -llx-26      x2-9a:+18  =  (s-13)(a;+2)      (a;  -6)  (a;  -3) 
:  ~ 


(x  +  3)  (*  -  5)  ~  x2  -  2x  -  15 

EXERCISES  85 

.,,,,.,    25  ,       48 

1.  Multiply^  by— 

2      25  5 

2.  Find  product  of  17X^X^-  Am.  7^- 

ou        Zu  lo 

.    45  ,       9  5 

3.  Divide  gj-  by  j^-  Ans.  ^« 

.    3w  .      c  m 

4.  Multiply—  by  —  Ans.  — 

CX              O  X 


300  PRACTICAL  MATHEMATICS 

.Safe     16c«x*      7d» 
5.  Find  product  of  —x-^X^,- 


6-  Divide  0,  ,  .  by 


x 


6+c 


u.i^xi±|. 


-n.          ..r 

ym     y 
13.  — S" 


IA    3x+y^4x 
14'  ~9~"  '  T 


16. 


?      2o>   ..llx'6  ...     22x' 


Sox      2c 

v/ A  <n  e       -. 

•       ~~  ^  /\  n      •  •*  **O.      ^^  „  . 

5ay     oy*  oay* 

.    2ax     5a»  10a»x 

O  NX. >1  TJJ» 

«•     «»       A»i      .  .!">. 


Arw. 


•  2n-l'n+l  '    2n-l 


2a+b     3a+2ft 

; sr"*"-: — r~r'  ^ITW. 


3a-26  '  4a+b  9a»-46» 

...     5y*    21c».35c^  3cy 

17.  Simplify  sriX-;  --  ^TTT^'  A«».  -r2 

J  7a3     4ax     7a'z  4a 


"5         x'  Ails. -.' 

x— 3  x— 4 

19.  Multiply —r     v~by^  ^.rw.  —  • 

x^ 


k    _.  .,    x^-Hx-lS,     x'-12x-45  x+1 

20-  Dlvlde  3.2_43._45-  by  iFZr6x^27'  ArW-  xT5' 

X          rrX        r»O              X              UX        £**  X  T^O 

~~  1  oX  ™f~  o(J        X    ~~\  OX  ~f~  «^\)  \    v/  -C    I    &  ,«             3J  "f"  1 

^^5^50"  '  ^-6^^7"j  Xx^l'  *'  x^l' 
a 


22.  Multiply  a2+2a6+62  by  ^^-2'  Am.  -^~~ 

11  a2 

23.  Multiply — r  by  a—  -r- 

1     1     6      a     b-a 
Suggestion.     — r  =  -r  — L  = — r- • 
a    o    ao    ao      ao 


aIo6—  a* 


.  .  ,     6  —  a  ,      ab—  a* 
Now  multiply  —  -r-  by  —  j-  —  • 


FRACTIONS  301 

24.  Multiply  £+-  by  a Ans.  °  7.    ' 

o     a  a  a,  b 

25.  Multiply  1  +- by  1  +r— : — •  A»s. 


Simplify  the  following: 


/     .     8x   \       /     .    2x  \ 
"    (V+^)  *  VT+x-3J 


42. 


x-y\ 


\x-2/     x+?//       \x-?/     X+T// 


26.  Multiply2+^byl-^|.  Ans. 


27.  Multiply  5-.,  by  -  +|-  •  Ans.  2. 

2x2+?/2       T/     2x 

28.  Multiply  —^rj-  by  -+£+-•  Ans.  a. 

ac+afe+&c       a     6     c 


00     c—  xvx3ax     a2  —  x2     a2+ox  3a; 

29.  -    —  X-rr-X~5  -  ,-J-r  —  rTT*  AfW.  -r— 
a  —  a;     4oy     c2—  a:2     bc+b.-v;  4i/ 

30.  -IK       3V  3, 


31      -  -r*yysy-y«     4a+a«     x2-?/2.  4+a  . 

Wi»     ^  \o^/         t         \o^  i  /N  o  ./I /to. 

(x-7/)2     (x+?/)2      x+a        ax?/2  xy+ay 


-  -.  s 

s-  XA 


a2—  x2      t  a—  x   .  o+x  .        x2—x 

.       ^       ;;  '.  ~r~     r  ~;^  r  "  A7  ~  " 

x2—  3x  —  4    x2  —  x    x—  4 


34 

x 


35    a2-a-6        5a+a2     .      a2  +2a 

•*•*•  o          0,1-       XN       o     i  •<  <-.     •  a  ^./-v  ^IftO.      i- 


a;2-a;-20^  a;2-a;-2    .    O-T-X 

ao.   —  prr      X    „    ,   <-,         F,  ~      :~  "^r~*  AW5.   3J. 


a2+6a-7      a&2+2a2b  .  a3-10a2+9a  .        ab-5b 

~~2a+b~     a2+a-42  :   a2-Tla+30*  '    a-9  ' 


a+b 


302  PRACTICAL  MATHEMATICS 

I         a    \  /     (z+y)«\  /      3m  \  3m 

»»•  I          ~  o  ,     J        I  A»*«.         ^— — •— — 


45.      _ 


46.    i_ 


47.  (x«-x  +  l)       ,+   +  l'  Xtw. 

48.  Simplify  ^  -  (^j  ?*  -ax(h  -*)  ?  and  get  ^(3A»* 


CHAPTER  XXVIII 
EQUATIONS  AND  FORMULAS 

247.  Subject  matter. — In  previous  chapters  a  number  of 
equations  were  presented  for  solution.     These  involved  simpli- 
fications of  various  kinds,  but  were  not  what  are  called  frac- 
tional equations;  that  is,  equations  in  which  fractions  occur. 
In  the  present  chapter,  besides  equations  like  those  previously 
considered,  will  appear  fractional  equations. 

248.  Order  of  procedure. — The  main  steps  in  the  solution 
of  a  simple  equation  have  already  been  given,  but  for  the  sake 
of  clearness  they  are  repeated  here. 

(1)  Simplify  the  equation;  that  is,  free  of  signs  of  grouping, 
perform  indicated  operations  of  multiplication  and  division  if 
possible,  clear  of  fractions,  etc. 

(2)  Transpose  all  the  terms  containing  the  unknown  to  the 
first  member  and  all  other  terms  to  the  second  member. 

(3)  Collect  terms. 

(4)  Divide  both  members  of  the  equation  by  the  coefficient  of 
the  unknown. 

(5)  Test  the  results  by  substituting  each  in  the  original  equa- 
tion. 

249.  Clearing  of  fractions. — A  fraction  is  an  indicated  divi- 
sion, and  usually  a  division  that  cannot  be  performed.     Hence 
when  an  equation  contains  fractions,  these  must  be  removed  by 
some  other  method  than  division. 

An  equation  can  be  cleared  of  fractions  by  multiplying  both 
members  of  the  equation  by  the  lowest  common  denominator  of  all 
the  fractions  in  the  equation. 

Example  1.     Solve  f+f  =  17  —  ^- 
o      8  1U 

'Ts        2C  3£ 

Solution.     (1)  Given  equation,  -+-  =  17  —  — • 

O      o  10 

(2)  Clearing  of  fractions  by  multiplying  each  term  by  the 
L.  C.  D . ,  40,  we  have,  Sx + 5x  =  680  -  4x. 

303 


304  PRACTICAL  MATHEMATICS 

(3)  Transposing,  8z+5z+4x  = 

(4)  Collecting  terms,  17z  =  680 

(5)  Dividing  by  coefficient  of  x,  x=  40. 

Test,  f+f-17-§  or  13-13. 

jf>  _  j  D 

Example  2.     In  the  equation  S  =  -^       r  solve  for  7. 

o.zzu 

PI  _   TO 

Solution.     (1)  Given  equation,  <Si  =  -7roo 


(2)  Clearing  of  fractions,  &220S  =  E-IR. 

(3)  Transposing,  IR  =  E-0.220S 

(4)  Dividing  by  coefficient  of  /,  7  =  — 

/v 

EXERCISES  86 

Solve  the  following  equations  for  the  unknown  letters,  and  test  the 
results. 

xx     10 


2x     7x     5x     x      4 
2'-3—  8+  18+24  =9 

??_Z?_11?     §5_u? 
'4      12  ~  36  ~  9  +2' 

2z     z     x     11 
*'  "5  +8  -4-40  =  0' 

x-fl.x+3 
5.  —  ~H  —  ~=2- 


.  4. 


2(g  +  l)     3(x+2) 
~~3~        "T~     !~~' 

2(z  +  D 

~~ 


3x+6 

Clearing  of  parentheses,  —  ^  ---  j  —  -~a~' 
o  4  O 

Clearing  of  fractions,  8x+8-9x-18     =2x+2. 
Transposing,  8x-9x-2x=  -8  +  18+2. 

Collecting  terms,  —  3x  =  12. 

Dividing  by  coefficient  of  x,  x  =  —4. 

8.     (2x  +  l)+2=(3x-2).  Ans. 


Q    x+1     x-1     3-x 
9"      5-     ~2~  !>— 

10.  |(x+2)  =  iU-3).  ^n«.  -12. 


EQUATIONS  AND  FORMULAS  305 


. 

11.  —j  ---  v  3       =3.  Ans.  -5. 

..    2-x     5x+21 

12.  —=  ---  =  —  =x+3.  Ans.  —  2Jf. 

2i  O 

13.  (z  +  l)2+2(z+3)2=3o:(z+2)+35.  Ans.  2. 

8  =  0.  Ans.  -Of,. 

15.  An,l. 


2 
16.  x-(3x-^^)  =  J(2z-57)-f.  4ns.  5. 


17.  %x+Q.25x-lx=x-3.  Ans.  12. 

0.36     0.09^-0.18 


679 


19.  Given  Sac  —  5c  =  17;  solve  for  c.  Ans.  ,. 

3a  —  5 

20.  Given  4a6c-56c  +  16=36c;  solve  (1)  for  a,   (2)  for  6,  (3)  for  c. 

_26c-4  4  4 

be  2c—ac'         2b—ab 

Solution  for  a.     (1)   Given  equation,  4abc  —  5fec  +  16  =36c 

(2)  Transposing,  4a6c  =  56c+36c  — 16. 

(3)  Collecting   terms,    4a6c  =  86c  — 16. 

8&C-16     26c-4 

(4)  Dividing  by  4oc,  a  =  — jr —      — j, 

Solution  for  b.     (5)  Transposing,   4a6c  —  5bc  —  3bc=  — 16. 

(6)  Collecting  terms,  (4ac  —  8c)6  =  — 16. 

—  16  4 

(7)  Dividing  by  4ac— 8c,  6=^ — ^-  =      _     • 

21.  Given  (b-x)(b+2x)  =62-2x2-36+4;  solve  (1)  for  b,  (2)  for  x. 

4  4-36 


22.  (a+b)x  +  (a-b)x  =  a*.  Ans.     • 

£ 

23.  |(a+.T)+l(2a+x)+i(3o+a!)=3a.  4ns.  a. 

24.  4(/+6+?/)+3(<+6  —  ?/)  =?/,  solvefor  /.  Am.    —b. 
The  following  are  formulas  that  occur  in  physics,  electricity,  etc. 

26.  Given  PD  =TFD1?  solve  for  P.  4ns. 


£> 

26.  Given  F=  —  —>  solve  for  TF,  £f,  r,  and  V. 

Fgr    WV*    WV2       fFJr 
Ans.  -£- ;  -fj—  ;  -r^—  :  *  '- 
V2'    fr    '    F^   ' 

27.  Given  /  =  „  ^     ?  solve  for  ^",  7?,  r,  and  n. 


20 


n(E-lr)    nE-IR      IR 
AnS-          n        '         I        '       In      'E-Ir 


306 


PRACTICAL  MATHEMATICS 
En 


Solution  for  E.     (1)  Given  equation,  7 


^  , 


(2)  Clearing  of  fractions,  IR+Inr=*En. 

IR+Inr_I(R+nr) 
n 


(3)  Dividing  by  coefficient  of  E,  E 

f» 

28.  Given  (1)  7-JL;  (2)  7-^-; and  (3) 


• ;  what  value  of 


n  will  make  (1),  (2),  and  (3)  identical?     Solve  for  n  in  (2)  and  (3). 


IR 


Ana.  1 ;  -_— 


Ir 


E-Jr'  E-IR 


29.  Given  7  =-H — >  solve  for  each  letter  in  terms  of  the  others. 
/     P     9 


p+q> 


L 


JP-. 
P-f 


Solution  for  p.     (I)  Given  equation,  7  =-+-• 

(2)  Multiplying  by  L.  C.  V.Jpq,  pq=fq+fp. 

(3)  Transposing,  pq— fp=fq. 

(4)  Collecting  terms,  (9—/)p— /?• 

(5)  Dividing  by  (q  -/),  P  =  ~Tf 

30.  If  T.  =-+£  +  /»  solve  for  k:  and  find  the  value  of  k  if  a  =  19,000, 


A-     a 
6  =  90,000,  and  Z=3180. 

31.  Given  —  = — h    »  solve  for  r. 

T       TI      TZ 

rl 


Ana. 


;2G40. 


32.  Given  E  =  RI+~,  find 
n 


33.  Given  nE  =  RI+~-,  find  7=- 
m  /t 


,£_ 
n     m 

34.  Given  Rt  =  Ri(l+al) ,  solve  for  R0,  a,  and  t 
in  terms  of  the  other  quantities. 

D  D  D  D  I? 

/t(        /O  —  *to     **i  —  *io 


Fio.  200. 


35.  Given  72  =  radius  of  a  circle,  h  =  height  of 
a  segment,  and  W  =  length  of  the  chord.  If  II' 
and  h  are  known,  find  R.  (See  Art.  120.) 

Solution.  In  the  figure,  AB  is  the  chord,  DC 
is  the  height,  and  AO  is  the  radius.  ADO  is  a 
right  triangle.  Hence, 

AO'-DO'-IB" 


or 


EQUATIONS  AND  FORMULAS  307 


Simplifying,  R*  -  (R*  -  2hR  +h2)  =  (~     . 

Simplifying,  R*-R*+2hR-h2  =  *• 


Transposing  and  collecting,  2A.R  =  (—  -  ]    +h2. 

\  2  / 


2A 

36.  When  a  bar  is  balanced  on  a  support  F  which  is  at  a  distance  a 
from  one  end  and  b  from  the  other,  with  weights  m  and  n  suspended 
from  either  end,  we  have  found  (see  Art.  68)  that  m:n  =  b:a.  This  gives 
the  equation,  am  =  bn.  A  lever  10  ft.  in  length  has  a  weight  of  1000  Ib.  on 
one  end.  Where  must  the  support  be  placed  so  that  a  weight  of  250  Ib. 
at  the  other  end  will  make  it  balance  ?  Disregard  the  weight  of  the  lever. 


FIG.  201. 

Solution.     Let  d  =the  distance  from  support  to  heavier  weight. 

Then  10  —  d  =  distance  from  support  to  lighter  weight. 

By  proportion,  250 : 1 000  =  d :  1 0  -  d. 

.:  2500-250d  =  1000d. 

Transposing,  -250d  -  lOOOd  =  -2500. 

Collecting  terms,  -  1250d  =  -  2500. 

Dividing,  d  =  2=number    of    feet    from    the 

1000-lb.  weight  to  the  support. 

The  weight  at  one  end  of  the  lever  in  Fig.  201  has  a  tendency  to  pull 
that  end  down,  that  is,  it  tends  to  make  the  lever  turn  about  the  fulcrum 
in  that  direction.  The  amount  of  this  turning  effect  is  known  as  a 
moment,  and  is  the  product  of  the  weight  by  the  length  of  its  lever  arm. 
Thus  am  is  the  moment  of  the  weight  m  acting  on  the  lever  arm  of 
length  a. 

Two  moments  that  tend  to  turn  in  opposite  directions  balance  each 
other  when  they  are  equal.  Two  such  moments  put  equal  to  each  other  give 
the  equation  used  in  solving  a  problem  in  levers. 

37.  If  a  lever  16  ft.  long  is  supported  at  a  point  18  in.  from  one  end, 
how  heavy  a  weight  can  a  man  weighing  150  Ib.  on  the  long  part  of  the 
lever  balance?     Disregard  the  weight  of  the  lever.  Ans.  1450  Ib. 

38.  Two  men  are  carrying  a  load,  weighing  350  Ib.,  on  a  pole  10  ft. 
long.     Find  the  weight  carried  by  each  if  the  weight  is  4  ft.  from  one  end 
of  the  pole.     Disregard  the  weight  of  pole.      Ans.  210  Ib.  and  140  Ib. 

39.  Place  a  fulcrum  under  a  lever  12  ft.  long  so  that  a  force  of  150  Ib. 


308 


PRACTICAL  MATHEMATICS 


at  one  end  will  balance  a  weight  of  1750  Ib.  at  the  other  end,  if  the  lever 
weighs  10  Ib.  per  foot.  Ans.  Nearly  1  ft.  3  in.  from  the  weight. 

When  the  weight  of  the  lever  is  taken  into  consideration,  the  moment 
of  each  arm  of  the  lever  is  the  weight  of  that  arm  multiplied  by  half  the 
length  of  the  arm.  Thus,  in  exercise  39: 

Let  x  «=the  length  of  the  short  arm  in  feet. 

Then  12— z=the  length  of  the  long  arm  in  feet. 

1750x  =  moment  due  to  weight. 
150(12— x)  =  moment  due  to  force. 

$i'10x  =  moment  due  to  short  arm. 
}( 12—  x)- 10(12—  z)  =  moment  due  to  long  arm. 

The  sum  of  the  moments  on  one  arm  is  equal  to  the  sum  on  the  other 
arm  if  the  lever  is  in  balance.  Then  the  equation  is 

1750x+5x*  =  150(12  -x)  +5(12  -x)!. 


. 


1            l 

iy         A 

1 

K*' 

,,     i             ,,                             '/T1  ' 
}<—  12-  >*<-  -12-->{ 

'>!<                                                  138  '                                                  > 

V 

0 

w 

FIG.  202. 

40.  In  the  arrangement  of  levers  for  a  platform  scale  shown  in  Fig. 
202,  find  the  lengths  of  the  arms  of  the  lever  resting  on  F4  so  that  a  weight 
of  1  Ib.  at  P  will  balance  a  load  of  1000  Ib.  at  W. 

Ans.  4  in.  and  40  in. 

In  the  following  exercises,  the  numbers  in  the  brackets  are  the  numbers 
of  the  formulas  as  given  in  previous  chapters,  where  their  meaning  can 
be  found. 


41.  [30]  A  =  Ihw.     Solve  for  w.  Ans.  w  = 

42.  [60]  V='2v*Rr'.     Solve  for  R.  Ans.  ^ 

43.  [46]  V=vR*h-TT*h.    Solve  for  r1  and  then  for  r. 

Ans. 


2h 
V 


44.  A  hollow  cylindrical  cast-iron  pillar  12  ft.  high  nnd  10  in.  in  outside 
diameter  is  to  weigh  1200  Ib.     Find  the  diameter  of  the  hollow. 

Ans.  7.7  in. 


EQUATIONS  AND  FORMULAS  309 

Suggestion.     Use  the  answer  of  exercise  43,  and  take  cast  iron  at  450  Ib. 
per  cubic  foot. 

2S 

45.  [47]  S  =  £/«.     Solve  for  s.  Ans.  — 

46.  [48]  T  =  \ps+A.     Solve  for  p.  Ans.  2(-T~A\ 

S 

47.  [49]-S  =  $(P+p)s.     Solve  for  p.  Ans.  2S~P*. 

S 


48.  [50]  T  =  $(P+j))s+B+b.     Solve  for  s.  Ans. 

49.  [66]  V  =  f  TiT3.    Solve  for  r.  Ans.   ., 

far-mx  FT2 

60.  Given  F  = TFZ-,  solve  for  x.  Ans.   —7-5- 


Ff  EP-p—Ff 

61.  Given  E  =  ,„      .  -,  solve  for  a;.  Ans.  — ^ — — • 

(P-ic)p'  £p 

rr'  Cr 

62.  Given  C =K — , — „  solve  for  r'.  Ans. 

r+r 

63.  Given  Q  =K- '       ,  solve  for  ti  and  a. 


r+r"  Kr  —  C 

a. 
aKtzT-dQ  dQ 


AnS"       aKT 
54.  Given  p 


55.  Given  //  =  1.600,000  -^—^(1+0.004:^,  solve  for 


g(6i+62)-l,600,000(6i-&i) 

6400(6!  -62) 

56.  Find  the  side  of  a  square  room  such  that  if  its  length  and  width 
are  increased  by  3  ft.,  the  area  is  increased  by  81  sq.  ft.         Ans.  12  ft. 

67.  Find  the  side  of  a  square  field  such  that  if  each  dimension  is  de- 
creased by  10  rods,  the  field  will  contain  400  sq.  rd.  less.         Ans.  25  rd. 

68.  A  rectangle  is  3  times  as  long  as  it  is  wide.     If  the  width  is  increased 
by  4  in.  and  the  length  decreased  by  5  in.,  the  area  is  increased  by  15  in.2 
Find  the  dimensions  of  the  rectangle.  Ans.  5  in.  by  15  in. 

59.  The  difference  between  the  base  and  the  altitude  of  a  triangle  is 
6  in.  and  their  sum  is  36  in.     Find  the  area  of  the  triangle. 

Ans.   157|  in.2 

60.  The  altitude  of  a  triangle  is  7  in.  longer  than  the  base.     If  the 
altitude  is  decreased  by  4  in.  and  the  base  increased  by  6  in.,  the  area  is 
increased  by  25  in.2     Find  the  altitude  and  base. 

Ans.  23  in.  and  16  in. 

61.  Find  the  number  such  that  we  can  take  ^  of  it  away  and  still  have 
i  of  320.  Ans.  576. 

62.  From  what  price  can  I  deduct  33^%  on  goods  which  cost  $3.20 
per  yard,  and  still  make  20%?  Ans.  $5.76. 


310  PRACTICAL  MATHEMATICS 

63.  Find  the  principal  that  invested  for  4}  years  at  3}%  per  annum 
will  give  an  amount  of  $604.50.  Ans.  $600. 

Solution.     Let  p  =  number  of  dollars  in  principal. 
Then  pX0.03$  X4}  =  0.1575p  =  number  of  dollars  in  interest, 
and         p+0. 1575p  =  number  of  dollars  in  amount. 
.'.p+0.1575p  =  694.50. 
1.1575p   =694.50 
p  =  600. 

64.  Find  the  rate  in  order  that  an  investment  of  $820  will  amount  to 
$912.25  in  3  yr.  9  mo.  Ana.  3%. 

66.  The  sum  of  $1100  is  invested,  part  at  5%  and  part  at  6%  per 
annum.  If  the  total  annual  income  is  $59,  how  much  is  invested  at 
each  rate?  Ans.  $700  and  $400. 

Suggestion.     Let  x  =  the  number  of  dollars  invested  at  5%, 
then  1100—  x  =  number  of  dollars  invested  at  6%. 
.•.0.05z+0.06(1100-z)  =59. 

66.  The  interest  on  $120  for  a  certain  time  at  6%  is  $16.56.     Find 
the  time.  Ans.  2  yr.  3  mo.  18  da. 

67.  The  interest  on  $584  for  2  yr.  8  mo.  7  da.  at  a  certain  rate  per  cent 
is  $94.121$.     Find  the  rate  per  cent.  Ans.  6%. 

68.  A  man  saves  $50  more  than  £  of  his  income,  spends  3  times  as 
much  for  living  expenses  as  he  saves,  and  spends  the  remainder  which 
is  $600  for  rent.     Find  his  income.  Ans.  $2400. 

69.  If  air  is  a  mixture  of  4  parts  nitrogen  to  1  part  of  oxygen,  how  many 
cubic  feet  of  each  are  there  in  a  room  40  ft.  by  30  ft.  and  12  ft.  high? 

Ans.  11,520;  2880. 

Suggestion.  Let  z=the  number  of  cubic  feet  of  oxygen  in  the  room. 
Then  4x  =the  number  of  cubic  feet  of  nitrogen  in  the  room. 

70.  A  room  is  -?$  as  wide  as  it  is  long.     If  the  length  were  decreased  by 
3  ft.  and  the  width  increased  by  3  ft.,  the  room  would  be  square.     Find 
the  dimensions  of  the  room.  Ans.  20  ft.  by  14  ft. 

71.  A  room  is  $  as  long  as  it  is  wide,  and  its  perimeter  is  70  ft.     Find 
the  area  of  the  room.  Ans.  300  ft.1 

72.  A  man  made  a  journey  of  560  miles,  part  at  the  rate  of  40  miles 
per  hour  and  part  at  50  miles  per  hour.     If  it  took  him  13  hours,  how 
far  did  he  go  at  each  rate?  Ans.  360  mi. ;  200  mi. 

73.  A  tree   189  ft.  tall  was  broken  into  two  pieces  by  falling.     If 
|  the  length  of  the  longer  piece  equals  }  the  length  of  the  shorter  piece, 
how  long  is  each  piece?  Ans.  100^  ft.;  88i?  ft. 

74.  Sirloin  steak  costs  1$  times  as  much  as  round  steak.     Find  the 
cost  of  each  per  pound  if  3  Ib.  of  sirloin  and  5  Ib.  of  round  cost  $2.28? 

Ans.  36  cents;  24  cents. 

75.  If  2  Ib.  of  butter  cost  as  much  as  5  Ib.  of  lard,  and  4}  Ib.  of  lard  and 
6  Ib.  of  butter  cost  $5.46,  find  the  cost  of  each  per  pound. 

Ans.  70  cents;  28  cents. 


EQUATIONS  AND  FORMULAS 


311 


250.  Thermometers. — Two  kinds  of  thermometers  are  in 
common  use.  The  Fahrenheit,  which  is  used  for  common 
purposes,  has  the  freezing  point  marked  32°,  and  the  boiling 
point  marked  212°.  The  centigrade,  which  is  used  for  all 
scientific  purposes,  has  the  freezing 
point  marked  0°  and  the  boiling 
point  marked  100°. 

It  is  seen  then  that  on  the  Fahren- 
heit scale  there  are  212° -32°  =  180° 
between  the  freezing  point  and  the 
boiling  point;  while  on  the  centi- 
grade scale  there  are  100°  in  the  same 
space.  Hence  180°  of  the  Fahren- 
heit scale  =  100°  of  the  centigrade 
scale.  These  relations  are  shown  in 
Fig.  203. 

EXERCISES  87 

1.  If  F  stands  for  the  number  of  degrees 
on    the    Fahrenheit   scale   and   C  for  the 
number  of  degrees  on  the  centigrade  scale, 
find  that 

(1)  C  =  §(F-32), 

(2)  F  =  esC+32. 

2.  A  temperature  of  176°  Fahrenheit  is 
what  temperature  centigrade?     [Use  (1)  of 
exercise  1.]  Ans.  80°. 

3.  A  reading  of  24°  centigrade  is  how 
many  degrees  Fahrenheit? 

Ans.  751°. 

4.  Given  that  the  following  metals  melt 
at   the  given  temperatures  in  Fahrenheit 

scale,  find  the  melting  point  of  each  in  the  centigrade  scale:  Wrought 
iron,  2822°;  steel,  2462°;  cast  iron,  2210°;  silver,  1832°;  lead,  620°;  tin, 
475°. 

Ans.  Wrought  iron,  1550°;  steel,  1350°;  cast  iron,  1210°;  silver,  1000°; 
lead,  326|°;  tin,  246J°. 

6.  60°  below  0°  Fahrenheit  is  what  on  the  centigrade  scale? 

Ans.  5H°  below  0°. 

6.  At  what  temperature  are  the  readings  on  the  two  thermometers 
the  same?  Ans.  40°  below  0°. 

Solution.     Let  x  =  the  reading  on  each  thermometer  scale.     Then  by 
the  formulas  of  exercise  1, 


212°F=  100°C 

-100*0 

ZI2-F210—  - 

BOILING          3 

200—  - 

• 

100  

! 

—  90 

180—  - 

; 

i 

—  80 

170  

5                        : 

160—  - 

I 

—  70 

150— 

nO- 

—  eo 

ISO- 

i 

^C— 

^ 

—   60 

110  — 

:                          ! 

100— 

: 

40 

T 

-  —   30 

70  — 

^ 

••  20 

60— 

60— 

10 

40  

32°  F«  0°C 

—     0*0 

82  F   30— 

FREEZING        - 

20  — 

\ 

'10  

> 

10 

0— 

20 

-10— 

-20— 

30 

-30— 

-40— 

, 

40 

FIG.  203. 


312  PRACTICAL  MATHEMATICS 

5(z-32)-!x+32. 
25x- 800*  Six +  1440. 
25x  -Six  -1440+800. 
-56x=2240. 
x=-40. 

.'.  the  reading  is  the  same  when  temperature  is  —40°. 
7.  With  the  oxyacetylene  process  of  welding,  the  temperature  of  the 
flame  is  sometimes  over  6000°  Fahrenheit.     What  is  this  on  the  centi- 
grade scale?  Ana.  Over  3315°. 

261.  Horse-power. — The  term  horse-power  was  first  used 
by  James  Watt,  the  inventor  of  the  steam  engine.  He  ascer- 
tained that  a  London  draught  horse  was  capable  of  doing  work, 
for  a  short  time,  equivalent  to  lifting  33,000  Ib.  1  ft.  high  in 
1  minute.  This  value  was  used  by  Watt  in  expressing  the 
power  of  his  engines,  and  has  since  been  universally  adopted 
in  mechanics. 

The  expression  foot-pounds  is  used  to  denote  the  unit  of  work; 
It  is  equivalent  to  a  force  of  1  Ib.  acting  through  a  distance  of  1  ft., 
or  a  force  of  ^  Ib.  acting  through  a  distance  of  2  ft.,  or  ^V  Ib. 
through  a  distance  of  10  ft.,  etc.  Horse-power  is  the  measure 
of  the  rate  at  which  work  is  performed.  One  horse-power  is 
equivalent  to  33,000  Ib.  lifted  1  ft.  in  1  minute,  or  1  Ib.  lifted 
550  ft.  in  1  second.  We  say  then  that  one  horse-power  equals 
33,000  foot-pounds  per  minute,  or  550  foot-pounds  per  second. 

Therefore,  the  horse-power  of  any  machine  can  be  found  by 
dividing  the  number  of  foot-pounds  of  work  done  in  1  minute 
by  33,000,  or 

Number  of  foot-pounds  of  work  per  minute 

Horse-power  =  -3pOO~ 

In  electric-power  machines  where  the  watt  is  used,  since  746 
watts  equal  1  horse-power,  we  have 

volts  X  amperes 

Horse-power  =  —   — =-.—       —  • 
746 

EXERCISES  88 

1.  What  horse-power  is  necessary  to  raise  a  block  of  stone,  weighing 
3  tons,  to  the  top  of  a  wall  40  ft.  high  in  2  minutes? 
Solution.     The  number  of  foot-pounds  per  minute  is 

6000X40 
— and  hence  the 

A 

6000X40 


EQUATIONS  AND  FORMULAS  313 

2.  What  horse-power  is  required  to  life  an  elevator,  weighing  4  tons, 
to  the  top  of  a  building  240  ft.  high  in  1J  minutes? 

Ans.  39  h.  p.  nearly. 

3.  What  horse-power  is  required  to  pump  30,000  barrels  of  water  per 
hour  to  a  height  of  45  ft.?     (Use  1  bbl.  =4.211  cu.  ft.) 

Ans.   179.4+  h.  p. 

4.  The  following  formula  gives  the  horse-power  of  a  steam  engine: 

PLAN 
33,000' 
where  H  =  indicated  horse-power, 

P  =  mean  effective  pressure  of  the  steam  in  pounds  per  square  inch, 

L  =  length  of  stroke  in  feet, 

A  =area  of  piston  in  square  inches,  and 

Ar=number  of  strokes  of  piston  (twice  the  number  of  revolutions) 

per  minute. 

Solve  the  formula  for  each  of  the  letters  P,  L,  A,  and  N  in  terms  of 
the  others. 

33,000//    .      33,000/7          33,000/7    .       33,000/7 

A  Yi  v       tf  —  - •    /     —  — i •     A    — /.... •     A/    —  ..       . 

'LAN    '       '    PAN    '       '    PLN  PL  A 

Remark.     It  is  to  be  noticed  that  the  formula  77  =  00  nnn  has,  in  the 

oOjUlHJ 

numerator,  an  expression  for  the  number  of  foot-pounds  per  minute. 
To  see  this,  note  that  PA  is  the  pressure  on  the  piston  in  pounds.  This 
times  L,  or  PL  A,  is  the  foot-pounds  for  one  stroke  of  the  piston.  Finally, 
multiplying  this  by  the  number  of  strokes  per  minute,  N,  gives  PLAN, 
the  number  of  foot-pounds  per  minute. 

6.  What  horse-power  will  a  steam  engine  with  a  cylinder  4  in.  in  diam- 
eter and  a  stroke  of  6  in.  develop  at  300  R.  P.  M.  of  the  crank,  if  the  mean 
effective  pressure  is  95  lb.?  Ans.  10.85+  h.  p. 

Suggestion.     Express  in  form  of  cancellation, 

4X4X0.7854X95X1X600 
33,000X2 

cancel  what  you  can  and  compute  by  slide  rule  if  you  wish. 

6.  Find  the  horse-power  of  an  engine  with  a  cylinder  10  in.  in  diameter, 
a  stroke  of  30  in.,  the  crank  making  96  R.  P.  M.,  and  the  mean  effective 
pressure  120  lb.  Ans.   137+  h.  p. 

7.  Find  the  diameter  of  a  cylinder  to  develop  95  h.  p.  with  a  stroke 
of  34  in.,  the  crank  making  110  R.  P.  M.,  the  boiler  pressure  being  80 
lb.,  and  the  mean  effective  pressure  65%  of  the  boiler  pressure. 

Ans.   11.1-  in. 

Suggestion.  Find  the  area  of  the  piston  by  the  formula  of  exercise  4, 
and  then  the  diameter  by  [27]. 

8.  An  engine  is  required  to  develop  50  h.  p.  with  an  average  effective 
pressure  of  46  lb.  on  a  piston  13  in.  in  diameter,  and  a  crank  shaft  speed 
of  100  R.  P.  M.     Find  the  length  of  the  stroke. 

Arts.   1.351+  ft.  or  1  ft.  4,3,;  in. 


314  PRACTICAL  MATHEMATICS 

9.  Find  the  mean  effective  pressure  on  the  piston  of  a  steam  engine 
with  a  cylinder  12  in.  in  diameter  and  a  piston  stroke  of  18  in.,  if  the  num- 
ber of  revolutions  is  110  per  minute  and  developed  horse-power  40. 

Ana.  35.37—  Ib.  per  square  inch. 

10.  In  gas  engines  it  is  not  easy  to  determine  the  mean  effective  pres- 

sure, so  the  formula  //  =  ™-^n  cannot  well  be  used.     As  a  result  of 

OO,UUU 

D*N 

experiments  with  engines  used  in  automobiles,  the  formula  H  •-s^r  is 

•.0 

often  used.     Here  D  is  the  diameter  of  the  cylinders  in  inches,  and  A'  is 
the  number  of  cylinders.     Find  //,  if  D  =4}  in.  and  N  =6.     Ans.  54.15. 

11.  In  "Locomotive  Data"  of  the  Baldwin  Locomotive  Works,  the 
following  formula  is  given  for  the  tractive  power  of  a  locomotive: 

C'SP 
~D~' 

where   C  =  diameter  of  cylinders  in  inches, 
S  =  stroke  of  piston  in  inches  , 
P=mean  effective  pressure  in  pounds  per  square  inch  =85%  of 

boiler  pressure, 

D=  diameter  of  driving  wheels  in  inches,  and 
T  =  tractive  power  in  pounds. 

Solve  for  each  of  the  letters  C,  S,  P,  and  D  in  terms  of  the  others. 

fDT    ~_DT   j>_DT    n 
'^~'  ' 


12.  Use  the  formula  of  exercise  11  and  find  T  when  C  =  16  in.,  5  =  22 
in.,  and  D  =  64  in.,  if  the  boiler  pressure  is  160  Ib.  per  square  inch. 

Am.  11,968  Ib. 

13.  To  find  the  pressure  on  a  lathe  tool  in  turning  steel  multiply  the 
area  in  square  inches  of  a  section  of  the  chip  cut  by  230,000.     In  cutting 
cast  iron  use  168,000. 

What  horse-power  does  it  take  to  turn  a  6-in.  steel  axle  making  30 
R.  P.  M.,  if  the  cut  is  3*5  in.  deep  with  a  feed  of  \  in.? 

Ana.  1.3  nearly. 

14.  The  power  that  is  transmitted  by  a  belt  depends  upon  the  pull  of 
the  belt  and  the  rate  at  which  it  travels.     The  power  may  be  given  as  a 
number  of  foot-pounds  per  minute  or  may  be  given  as  horse-power. 
Different  makers  of  belts  and  writers  on  the  subject  give  different  values 
as  the  working  strength  for  belts.     The  allowed  pull  for  single  belts  is  from 
30  to  60  Ib.  per  inch  of  width  of  belt.     For  double  belts  it  is  from  60  to 
100  Ib.  per  inch  of  width. 

Show  that  the  following  formula  gives  the  horse-power  transmitted  by 
a  belt: 

FWS 
33,000' 
where    //  =  horse-power, 

F  =  pull  in  pounds  per  inch  of  width, 
W  =  width  in  inches, 
5  =  speed  of  belt  in  feet  per  minute. 


EQUATIONS  AND  FORMULAS 


315 


Solve  for  each  letter  involved  in  the  formula. 

15.  Find  the  horse-power  that  can  be  transmitted  by  a  belt  14  in. 
wide,  if  the  pull  allowed  per  inch  of  width  is  90  Ib.  and  the  speed  is  5000 
ft.  per  minute.  Ans.  190fr. 

16.  Find  the  horse-power  transmitted  by  a  single  belt  6  in.  wide, 
running  over  a  pulley  16  in.  in  diameter,  and  making  350  R.  P.  M.,  if  the 
pull  of  the  belt  is  45  Ib.  per  inch  of  width.     Allow  2%  for  slipping. 

Ans.  11.85. 

17.  If  a  single  belt  6  in.  wide  transmits  7  horse-power,  find  its  speed 
per  minute,  allowing  35  Ib.  pull  per  inch  of  width. 

Ans.  1100  ft.  per  minute. 

18.  Find  the  horse-power  transmitted  by  a  10-in.  double  belt,  run- 
ning over  a  36-in.  pulley,  making  420  R.  P.  M.     Use  a  pull  of  75  Ib. 
per  inch  of  width.  Ans.  89.96. 

19.  Assuming  a  tension  of  50  Ib.  per  inch  of  width  for  a  single  belt, 
and  using  D  for  diameter  of  pulley  in  inches,  W  for  width  of  belt  in  inches, 
R  for  the  number  of  revolutions  per  minute,  and  H  for  the  horse-power 
transmitted,  then 

rr^DRW 

~  2520 ' 

Use  ir=?r  and  derive  this  formula.     Solve  for  W,  D,  and  R  in  terms  of 
the  other  letters. 

Remark.  In  the  above  formula  it  is  assumed  that  the  pulleys  are  prac- 
tically of  the  same  size  so  that  the  arc  of  contact  is  180°.  If  the  pulleys 
differ  in  diameter  the  arc  of  contact  of  the  belt  on  the  smaller  pulley  should 
be  found  and  the  following  table  used  in  finding  the  horse-power  that  can 
be  transmitted. 


If  angle  of  con- 
tact is  . 

90° 

100° 

110° 

120° 

130° 

140° 

150° 

160° 

170° 

Multiply  by  

0.65 

0.70 

0.75 

0.79 

0.83 

0.87 

0.91 

0.94 

0.97 

20.  Using  the  notation  of  the  preceding  exercise  and  80  Ib.  for  the 
tension  per  inch  of  width  for  double  belts, 

rj_DRW 

~  1575  ' 

Derive  this  formula  if  ir=V-     Solve  for  D,  R,  and  W  in  terms  of  the 
other  letters. 

21.  Find  the  width  of  a  single  belt  to  transmit  3  horse-power  when 
running  over  a  pulley  15  in.  in  diameter,  making  220  R.  P.  M.     Allow  a 
pull  of  50  Ib.  per  inch  of  width.  Ans.  2.3  in. 

22.  Find  the  number  of  R.  P.  M.  a  pulley  4  ft.  in  diameter  must  make 
that  transmits  120  horse-power  through  a  double  belt  10  in.  wide,  having 
a  pull  of  80  Ib.  per  inch  of  width.  Ans.  393  J. 


316  PRACTICAL  MATHEMATICS 

23.  How  much  work  is  done  in  lifting  150  Ib.  from  a  mine  1100  ft. 
•loop?  How  many  horse-power  would  it  take  to  lift  this  weight  from  the 
mine  in  1  !•  minutes?  Ana.  3J  horse-power. 

252.  Relation  of  resistance,  electro-motive  force,  and 
current. — Resistance,  R,  is  measured  in  ohms ;  voltage  or  elec- 
tro-motive force,  E.  M.  F.  or  E,  in  volts;  and  current,  7,  in 
amperes.  The  law  connecting  these  is  stated  as  follows: 

volts        .     E 

Amperes  =  -; or  /  =  -=• 

ohms  R 

This  law  is  Ohm's  Law,  and  is  the  fundamental  law  for 
electrical  work.  It  is  an  algebraic  equation,  and  any  one  of  the 
numbers  can  be  found  if  the  other  two  are  given. 

Exercises  27,  28,  32,  and  33,  pages  305  and  306,  are  forms  of 
this  equation. 

EXERCISES  89 

E  E 

1.  Solve  the  equation  J  =  p  for  E  and  R.  Ans.  E  =  IR;  R**j- 

2.  In  a  certain  circuit  the  voltage  is  measured  and  found  to  be  1.5 
volts.     If  the  total  resistance  is  12  ohms,  what  is  the  current  in  amperes? 

Ans.  0.125  ampere. 

3.  Find  the  number  of  amperes  of  current  sent  through  a  circuit  of 
20  ohms  resistance  by  one  Daniell's  cell  which  has  an  E.  M.  F.  of  1.03 
volts.  Ans.  0.0515. 

4.  Find  the  strength  of  current  from  50  Daniell's  cells  united  in  series, 
assuming  the  E.  M.  F.  of  each  cell  to  be  1.03  volts,  the  resistance  within 
each  cell  0.3  ohm,  and  the  external  resistance  25  ohms. 

Ans.  1.2875  amperes. 

Remark.  The  student  not  familiar  with  the  meaning  of  "cells  united 
in  series"  can  note  the  fact  that  50  cells  in  series  have  50  times  the 
E.  M.  F.  of  a  single  cell,  and  the  resistance  within  the  cells  is  50  times 
that  of  one  cell.  The  formula  of  exercise  27,  page  305,  can  be  used. 

6.  An  electric  bell  has  a  resistance  of  450  ohms  and  will  not  ring  with  a 
current  of  less  than  0.06  ampere.  Neglecting  battery  and  line  resistance, 
what  is  the  smallest  E.  M.  F.  that  will  ring  the  bell?  Ans.  27  volts. 

6.  If  an  electric  car  heater  is  supplied  with  500  volts  from  the  trolley, 
how  great  must  its  resistance  be  so  that  the  current  may  not  exceed 
2.5  amperes?  Ans.  200  ohms. 

7.  A  certain  wire  has  a  resistance  of  1  ohm  for  every  30  ft.  of  its  length. 
What  must  be  the  E.  M.  F.  in  order  that  a  current  of  0.4  ampere  may 
flow  through  1  mile  of  the  wire?  Ans.  70.4  volts. 


EQUATIONS  AND  FORMULAS  317 

253.   Resistance    of    conductors.  —  Consider    the    formula, 

K  ==     K, 

a 

where  R  is  resistance  in  ohms,  I  the  length  of  the  conductor  in 
some  unit,  a  the  area  of  the  cross  section  of  the  conductor  in 
some  unit,  and  k  a  constant  value  depending  upon  the  material 
in  the  conductor  and  upon  the  units  of  length  and  cross  section 
used.  This  formula  is  the  fundamental  formula  in  wiring 
calculations. 

In  engineering  practice,  the  length  I  is  taken  in  feet;  the 
area  a  is  taken  in  circular  mils,  which  equals  the  square  of  the 
diameter  in  thousandths  of  an  inch  when  the  conductor  is 
circular  (see  Art.  132)  ;  and  k  is  the  resistance  in  ohms  of  1  ft. 
length  of  the  conductor  having  a  cross  section  of  one  circular  mil. 
That  is,  k  is  the  mil-foot  resistance  of  the  conductor.  The 
formula  can  then  be  written 


=  -. 

a2 

If  then  it  is  required  to  find  the  resistance  of  any  length  of 
any  size  wire  of  any  material,  it  is  only  necessary  to  know  the 
mil-foot  resistance  for  the  -material,  and  substitute  the  values 
in  the  formula. 

The  following  table  gives  the  mil-foot  resistances  for  the 
materials  named  at  zero  degrees  centigrade  : 

Aluminum  ..........................  17  .  5  ohms. 

Copper  (commercial)  .................  9.6  ohms. 

German  silver  .......................  125  .  7  ohms. 

Iron  (pure)  .........................  58.  3  ohms. 

Iron  (telegraph  wire)  .................  90  .  0  ohms. 

Platinum  ...........................  54  .  3  ohms. 

Silver  ..............................  9.1  ohms. 

Zinc  ...............................  33.8  ohms. 

Example  1.  What  is  the  resistance  of  2  miles  of  No.  12, 
B.  and  S.  gage  commercial  copper  wire? 

Solution.     No.  12,  B.  and  S.  gage  has  an  area  of  6530  C.  M. 

2  miles  =2X5280  ft. 

2X5280X9.6 
,.R  =  —    -^-TC.?;—  —==15.  5  ohms.  Ans. 

DOOU 


318  PRACTICAL  MATHEMATICS 

If  it  is  required  to  determine  the  size  of  wire  of  a  given  length 
to  have  a  given  resistance,  the  formula  is  solved  for  d*.     This 

M    lk 
gives  d2  =  ^- 

Example  2.    Find  the  B.  and  S.  gage  of  pure  iron  wire  to  have 
a  resistance  of  3  ohms  per  mile. 

Solution.     Substituting  in  the  formula, 

d* - 528°X58.3 _  102)60g  c  M 

o 

d  =  V102.608  =  320.3  mils  =  0.3203  in. 

The  nearest  to  this  size  is  No.  0,  B.  and  S.  which  is  0.3249  in. 
in  diameter. 

If  it  is  required  to  find  the  length  of  wire  of  a  given  size  to 
have  a  given  resistance,  the  formula  is  solved  for  I.     This  gives 

Ra 

=  *' 

Example  3.     Find  the  length  of  a  No.  20,  B.  and  S.  gage  silver 
wire  to  have  a  resistance  of  5  ohms. 

Solution.     No.  20,  B.  and  S.  has  an  area  of  1022  C.  M. 
5X1022 


9.1 


=  561. 5  ft.  nearly.  Ans. 


EXERCISES  90 


1.  Find  the  resistance  of  340  ft.  of  No.  25,  B.  and  S.  gage  German 
silver  wire.  Ans.   133.4—  ohms. 

2.  Find  the  resistance  of  20  miles  of  trolley  wire,  made  of  commercial 
copper,  and  No.  00,  B.  and  S.  Ans.  7.6  +  ohms. 

3.  Find  the  resistance  of  10  miles  of  the  "third  rail"  conductor  on  an 
elevated  railroad.     The  "third  rail"  is  iron  and  has  a  cross  section  of 
5.88  in.1  Ans.  0.63  ohm  nearly. 

4.  Find  the  resistance  of  500  turns  of  No.  30,  B.  and  S.  gage  silver 
wire,  about  a  core  j  in.  in  diameter.  Ans.  8.69+  ohms. 

5.  Find  the  B.  and  S.  gage  of  commercial  copper  wire  to  give  a  resist- 
ance of  40  ohms  per  mile.  Ans.  No.  19  nearly. 

6.  Find  the  length  of  No.  16,  B.  and  S.  gage  commercial  copper  wire  to 
have  a  resistance  of  20  ohms.  Ans.  5381  ft. 

7.  The  resistance  of  a  certain  commercial  copper  wire,  1  ft.  long  and 
1  C.  M.  in  cross  section,  is  10.79  ohms.     A  wire,  525  ft.  long,  has  a  cross 
section  of  4117  C.  M. ;  what  is  its  resistance?  Ans.  1.376  ohms. 


CHAPTER  XXIX 

EQUATIONS  WITH  MORE  THAN  ONE 
UNKNOWN 

254.  In  previous  chapters  the  equations  solved  have  in- 
volved only  one  unknown  letter.     In  the  present  chapter  will 
be  considered  equations  involving  more  than  one  unknown 
letter.     These  letters  will  occur  to  the  first  power  only,  or  if 
squared,  will  be  such  as  are  easily  handled. 

255.  Indeterminate  equations. — If  we  have  the  equation 
x  —  y  =  2,  and  try  to  solve  it  for  the  unknown  numbers  x  and  y, 
it  is  evident  that  we  can  get  numerous  pairs  of  values  for  these 
numbers. 

Thus,  z  =  3  and  y  =  l  is  a  pair  of  values  that  satisfy  the  equation;  so 
are  z  =  4  and  y  =  2;  x  =  5  and  ?/=3;  x  =  10  and  y=8. 

Such  an  equation  is  called  an  indeterminate  equation. 
Find  pairs  of  values  that  satisfy  the  following: 

1.  x+y  =  lQ.          2.  x-y  =  lQ.          3.  2x+y  =  20. 
4.  3z+2t/  =  40.       5.  7x-8y  =  27.      6.  3x-8y  =  lQ. 
1.6x-2y=-2.     B.4x-8y=-lO. 

256.  Simultaneous  equations. — Take  the  two  indeterminate 
equations 

(1)  x  —  y  =  2,  and 

(2)  x+y  =  12. 

Pairs  of  values  that  satisfy  (1)  are  x  =  3,  y  =  l;  #=4,  y  =  2; 
x  =  5,  y  =  3;  x  =  Q,  y  =  4;  x  =  7,  y  =  5;  etc.  Pairs  of  values  that 
satisfy  (2)  are  x  =  2,  7/  =  10;  x  =  3,  y  =  9;  x  =  Q,  y  =  6',  x  =  7, 
y  =  5;x  =  8,  y  =  4;etc. 

It  is  noticed  that  the  pair  of  values,  x  =  7  and  y  =  5,  is  found 
in  both  sets  of  values;  that  is,  these  values  satisfy  both  equa- 
tions. Such  a  set  of  two  equations  satisfied  by  a  pair  of 
values  for  the  unknown  is  called  a  system  of  simultaneous 
equations. 

319 


320  PRACTICAL  MATHEMATICS 

If,  as  in  the  system  just  considered,  there  is  only  one  pair 
of  values  that  satisfy  both  equations,  the  equations  are  called 
independent. 

If  the  equations  are  as  in  the  set  x+2y  =  7  and  3x+6y  =  15, 
where  there  is  no  pair  of  values  alike  for  both,  the  equations 
are  called  inconsistent. 

If  the  equations  are  as  in  the  set  2x+y  =  10  and  6z  =  30  —  3y, 
where  every  pair  that  satisfies  one  will  satisfy  the  other,  the 
equations  are  called  equivalent. 

257.  Solution  of  independent  equations. — It  is  evidently 
not  convenient  to  find  a  pair  of  values  that  will  satisfy  two  inde- 
pendent equations,  by  writing  down  pairs  of  values  for  each 
equation,  and  then  selecting  the  pair  which  is  the  same  in  each. 
It  remains  then  to  devise  a  way  that  is  shorter. 

The  different  methods  of  solution  that  have  been  devised 
are  alike  in  that  an  equation  is  obtained  that  has  only  one 
unknown.  We  say  then  that  one  unknown  has  been  eliminated. 
There  are  three  ways  of  eliminating  an  unknown.  They  are: 

(1)  Elimination  by  adding  or  subtracting. 

(2)  Elimination  by  substitution. 

(3)  Elimination  by  comparison. 

258.  Elimination  by  adding  or  subtracting.- 

Example  1.     Solve  x— y  =  2  and  x+y  =  12  for  x  and  y. 
Solution.     (1)  x  —  y  =  2. 

(2)  z-h/  =  12. 

Add  equations  (1)  and  (2),  first  member  to  first  member  and 
second  member  to  second  member,  and  we  have 

(3)  2z=14. 
.'.(4)  z  =  7. 

Substituting  this  value  of  x  in  (1)  gives 

(5)  7-y  =  2. 
.'.   (6)  y  =  5. 

Hence  the  pair  of  values  that  will  satisfy  both  equations  is 
x  =  7  and  y  =  5. 

Example  2.     Solve  3z+2y  =  21  and  7x  —  5v/  =  20  for  JT  and  y. 
Solution.     (1)  3x+2y  =  2l. 
(2)  7z-5t/  =  20. 

Multiplying  (1)  by  7  gives     (3)  21x+14j/=  147. 
Multiplying  (2)  by  3  gives     (4)  21z-15y  =  60. 


EQUATIONS  WITH  MORE  THAN  ONE  UNKNOWN    321 

Subtracting  (4)  from  (3)  =     (5)  29y  =  87. 

'.  (6)  y-3. 

Substituting  in  (1)  gives           (7)  3s +6  =  21. 

.'.  (8)  3^  =  21-6. 

(9)  3z  =  15. 

(10)  a;  =  5. 
Hence  x  =  5  and  y  =  3  are  the  required  values. 

This  way  of  eliminating  one  of  the  unknown  numbers  is 
called  elimination  by  adding  or  subtracting.  The  following 
rule  may  be  given  for  the  process: 

RULE.  (1)  Multiply  each  equation,  if  necessary,  by  such  a 
number  as  will  make  the  absolute  values  of  the  coefficients  of  one 
of  the  unknown  numbers  the  same  in  both  of  the  resulting  equations. 

(2)  Add  or  subtract  the  corresponding  members  of  the  resulting 
equations  so  as  to  eliminate  the  unknown  number  having  coefficients 
equal  in  absolute  value. 

It  should  be  noticed  that  we  add  when  the  coefficients  are 
of  opposite  signs,  and  subtract  when  they  are  of  like  signs. 

259.  Elimination  by  substitution. — The  elimination  can  often 
be  performed  more  easily  by  using  a  method  called  substitution. 
Consider  the  following  system  of  equations: 
y  =  42  —  7x  and  3x  —  y  =  8. 

Solution.     (1)  y  =  42  —  7x. 

(2)  3x-y  =  8. 

Substituting  the  value  of  y  from  (1)  into  (2)  gives 

(3)  3z-(42-7z)=8. 

(4)  Zx-42+7x  =  8. 

(5)  3z+7z  =  42+8. 

(6)  10s  =  50. 

(7)  a;  =  5. 
Substituting  in  (1)  gives  (8)  y  =  42-35. 

.'.  (9)y  =  7. 
Hence  x  =  5  and  y  =  7  are  the  values. 

This  method  of  eliminating  is  called  elimination  by  sub- 
stitution, and  can  generally  be  used  to  good  advantage  when 
one  equation  is  much  simpler  in  form  than  the  other.  The 
following  rule  may  be  given  for  the  process: 

RULE.  Solve  one  of  the  equations  for  the  value  of  one  of  the 
unknown  numbers,  and  substitute  this  value  in  place  of  that 

21 


322  PRACTICAL  MATHEMATICS 

number  in  the  other  equation.     This  will  give  an  equation  with 
but  one  unknown  number. 

260.  Elimination  by  comparison.— 
Example.     Solve  z+4y  =  21  and  Zx  —  y=ll. 
Solution.       (1)  x+4y  =  21. 

(2)  3z-2/  =  ll. 
Solving  (1)  for  x  gives  (3)  x  =  2l—4y. 

Solving  (2)  for  x  gives  (4)  a?  = — =-*• 

o 

By  axiom  (5),  equations  (3)  and  (4)  give 

(5)     21-4y-ii±£ 

Clearing  of  fractions,         (6)  63-12i/  =  ll+y. 

(7)-12y-y-ll-63. 

(8)  -13y=-52. 

(9)  y-4. 
Substituting  in  (3)  gives  (10)  x  =  21  - 16. 

(11)  z  =  5. 

Hence  x  =  5  and  y  —  4  are  the  values. 

This  method  of  elimination  is  called  elimination  by  com- 
parison. The  following  rule  may  be  given  for  the  process: 

RULE.  Solve  each  of  the  equations  for  the  value  of  one  of  the 
unknowns,  and  equate  these  values.  This  forms  an  equation 
having  but  one  unknown. 

261.  Suggestions. — Any   method   of   elimination   may   be 
used.     Usually  one  of  the  methods  is  shorter  than  the  others. 
Elimination  by  adding  or  subtracting  can  usually  be  used  to 
the  best  advantage. 

Free  the  equations  of  signs  of  grouping  before  eliminating. 
Usually  clear  of  fractions  before  eliminating. 

Much  practice  and  dealing  with  examples  will  help  one  to 
determine  what  method  to  use. 

EXERCISES  91 

In  exercises  1  to  10  eliminate  in  each  by  all  three  methods. 

1.  z+y**8  and  x— y=2.  Ana.  z  =  5,  y=3. 

2.  2x+y  =  10andz+2j/  =  ll.  Am.  z=3,  y  =  4. 

3.  4x-3y  =  8  and  2z+y  =  14.  Ana.  x-5,  y**4. 

4.  3x+2y=26and  5x-2y  =  38.  Ana.  x«=8,y  =  l. 

5.  7x+y=42  and  3x-y=8.  Ana.  z«5,  y-7. 


EQUATIONS  WITH  MORE  THAN  ONE  UNKNOWN    323 

6.  8x+6y  =  10  and  5x+2y  =  1.  Ans.  x=—l,y  =  3. 

7.  7x  — 9y  =  13  and  5x+2?/  =  10.  Ans.  x  =  l*|,  2/=A- 

8.  3x+5y=8  and  2x  —  3y  =  12.  Ans.  x=4T85,  y=  —  f§. 
Solution.     (1)  3x  +  5y  =  8. 

(2)  2x-3y  =  l2. 

Multiplying  (1)  by  2  gives  (3)  6x  +  10?/  =  16. 
Multiplying  (2)  by  3  gives  (4)  6x- 9^  =  36. 
Subtracting  (4)  from  (3)  gives  (5)  19?/=  -20. 

Substituting  in  (1),     x  =  4j\,. 

9.  5x+6y  =  17  and  6x+5?/  =  16.  Ans.  x  =  l,  y=2. 
10.  x-lly  =  l  and  llly-9x  =  99.                          Ans.  a;  =  100,  ?/  =  9. 
In  exercises   11  to  14  draw  the  graphs  of  the  equations,  determine 

values  from  these,  and  solve  equations  by  some  method  of  elimination. 
(See    Chapter    XXXIII.) 

j.1.  x~r~?/  =  17  niici  x  —  y==i.  Ans.  x  —  1^,  t/===o. 

12.  3x+4y  =  24  and  5x  —  Qy  =  2.  Ans.  x  =  4,y  =  3. 

13.  5x+9?/  =  28and  7x+3y  =  20.  Ans.  x  =  2,  y  =  2. 

14.  5x-2y  =  l,  and  4x+5?/=47.  Ans.  x  =  3,  y  =  7. 
Solve  the  following  by  some  method  of  elimination : 

:c-j-l     y-\-2       .  x-\-y    y~\~2  .  _ 

Ans.  x=  -2,  y=-2. 

?_^_A    ^  ?_^_1 

3     6~2a!      5~10~2' 

18.  ^+|  =  5  and  a-6  =  4.  Ans.  a  =  10,  6  =  6. 

O         — • 

<0    n  +  1     3m  — 5        ,  n  +  1     n  —  m 

19.  ^Q-= — 2 —          Tb~=~~8 —  Ans.  m=3,n  =  lQ. 


f\f\           X              £               U        \      LJ               ~                       ,       *J**              U               J.  A               **M                f~  f,                            f* 

•jfl       —  1 1    Q  n  fl  A  *n  a      v  —  *^     II  —  s 

a\l.         „                  —       —  \J  tillU          j.  _           —  V.                        yi  Jto.  X  —  v»,   y  —  a, 

21.  -—  ~  =4  and  =+-~^=3.  Ans.  x  =  14,  y  =  15. 

ft-xwl        .XT/!  o+c              d  — 6 

22.-+f=-rand — --,  =  -—,•  Ans.x=— 7-7^-.  2/=— rnr- 

a     b     ab           c     d     cd  ad+bc          ad+bc 

Solution.      (1)  -+r=~T' 
06     ab 


Clearing  of  fractions,  (3)  bx+ay  =  l. 
(4)  dx—cy  =  l. 

Multiplying  (3)  by  c,  (.5)  lcx+acy=c. 
Multiplying  (4)  by  a,  (6)  adx  —  acy  =  a. 
Adding  (5)  and  (6),  (7)  (ad+bc)x  =  a+c. 

•    •      -  a+c 


324  PRACTICAL  MATHEMATICS 

Here  it  is  beat  to  solve  for  y  by  eliminating  x  rather  than  to  substitute 
the  value  of  y  in  (1)  or  (2). 

23.  The  sum  of  two  numbers  is  15  and  their  difference  is  1.     What  are 
the  numbers?  Ana.  7  and  8. 

24.  Three  times  one  number  plus  four  times  another  number  is  ten, 
and  four  times  the  first  plus  the  second  is  nine.     What  are  the  two 
numbers?  An*.  2  and  1. 

25.  What  is  the  fraction  which  equals  |  when   1  is  added  to  the 
numerator,  but  equals  i  when  1  is  added  to  the  denominator? 

Ana.  iV 

26.  Given  *S  =  — ry  and  T=  jr, eliminate  N  and  find  the  value  of   T 

j-vrir 

in  terms  of  the  remaining  letters.  Ana.  T '  =    -_»  • 

Suggestion.     Solve  each  equation  for  N  and  eliminate  by  comparison. 

27.  The  formulas  of  exercise  26  are  used  in  lathe  work.     In  these 
formulas,  T  is  the  time  in  minutes;  S  is  the  cutting  speed  in  feet  per 
minute;  D  is  the  diameter  of  the  work  in  inches;  N  is  the  number  of 
revolutions  per  minute;  L  is  the  length  of  the  part  to  be  turned  in  inches; 
F  is  the  feed  and  is  expressed  as  the  numbers  of  turns  to  give  a  sidewise 

movement  of  1  in.  Thus,  a  feed  of  16  means 
that  each  cut  is  ^  in.  wide.  Find  the  time  to 
turn  a  piece  3  in.  in  diameter  and  2  ft.  long, 
if  the  feed  is  20  and  the  cutting  speed  15  ft. 
per  minute.  Ans.  25.13+  min. 

28.  In  locating  and  boring  holes  in  a  drill 
jig,  it  is  necessary  to  find  the  diameters  of 
three  circular  disks  tangent  two  and  two, 
whose  centers  are  at  distances  of  0.765  in., 
0.710  in.,  and  0.850  in.  Find  the  diameters 
of  the  three  circles. 
FJG  204  Solution.  The  disks  are  placed  as  shown 

in  Fig.  204. 

Let  x,  y,  and  z  =  the  radii  of  the  circles  centered  at  A,  B,  and  C  respec- 
tively. 

Then  (1)  z+2/  =  0.850. 

(2)  x+z  =  0.710. 

(3)  y+z  =  0.765. 

Subtracting  (2)  from  (1)  =  (4)  y—z  =0.140. 
Adding  (3)  and  (4)  =  (5)      2y  =  0.905. 

(6)       77=0.4525. 
Substituting  value  of  y  in  (1)  gives  x  +0.4525  =0.850. 

/.  i  =  0.3975. 

Substituting  value  of  y  in  (3)  gives  0.4525+2  =  0.765. 

/.  z=0.3125. 

The  diameters  of  the  circles  are:  at  A,  0.795  in.;  at  B,  0.905  in.;  at  C, 
0.625  in. 


EQUATIONS  WITH  MORE  THAN  ONE  UNKNOWN    325 

29.  As  in  the  last  exercise,  three  holes  are  to  be  bored,  the  distance 
between  whose  centers  shall  be  0.650  in.,  0.790  in.,  and  0.865  in.  respec- 
tively.    Find  the  diameters  of  the  required  disks. 

Ans.  0.725  in.;  0.575  in.;  1.005  in. 

30.  Three  points  A,  B,  and  C  are  located  as  shown  in  Fig.  205.     Three 
disks  are  centered  at  these  points  and  tangent  two  and  two.     Find  the 
diameters  of  the  disks. 

Ans.  At  A,  0.8960  in.;  at  B,  0.9716  in.;  at  C,  0.9700  in. 
Suggestion.     Let  x,  y,  and  2=  radii  of  circles  centered  at  A,  B,  and  C 
respectively. 

Then  x+y=  v/0.6802+0.6402. 
x+2  =  \/0.8802+0.3102. 


FIG.  205. 


31.  From    geometry    we    know 
that  the  sum  of  the  three  angles 
of    a   triangle   is   180°.     Find  the 
three   angles   of  a  triangle  if  the 
sum    of    twice    the   first   and  the 
second  is  90°  more  than  the  third, 
and  the  sum  of  the  first  and  twice 
the  third  is  70°  more  than  twice 
the  second.      Ans.  50°;  60°;  70°. 

32.  A  man    has    $98    in    dollar 
bills,    half-dollars,    and    quarters. 
Half  of  the  dollars  and  f  the  half- 

dollars  are  worth  $31,  and  f  the   half-dollars  and  -|  the  quarters  are 
worth  $10.     How  many  pieces  has  he  of  each? 

Ans.  48  dollars;  70  half-dollars;  60  quarters. 

33.  In  a  factory  where  700  men  and  women  are  employed,  the  average 
daily  pay  for  the  men  is  $4.25,  and  for  the  women  $2.25.      If  $2415  is 
paid  daily  for  labor,  find  the  number  of  men  and  the  number  of  women 
employed.  Ans.  420  men  ;  280  women. 

34.  A  lever  is  balanced  on  a  fulcrum  with  a  weight  of  40  Ib.  at  one  end 
and  50  Ib.  at  the  other.     If  5  Ib.  is  added  to  the  40-lb.  weight,  the  50-lb. 
weight  will  have  to  be  placed  1  ft.  farther  from  the  fulcrum  to  balance 
the  lever.     Find  the  lengths  of  the  two  arms  of  the  lever  at  first. 

Ans.  10  ft.  ;  and  8  ft. 

Suggestion.     Let  z  =  the  length  in  feet  of  the  long  arm, 
and  ?/=the  length  in  feet  of  the  short  arm. 
Then  40z  =  507/, 


36.  A  lever  is  balanced  on  a  fulcrum  with  a  weight  Wi  on  one  arm  and 
Wi  on  the  other.  If  p  pounds  are  added  to  w\,  w^  has  to  be  moved  over 
/  feet  to  balance  the  lever.  Find  the  lengths  of  the  two  arms  of  the 


lever. 


Ans.  - — •  and 


326  PRACTICAL  MATHEMATICS 

36.  A  beam  that  is  supported  at  its  ends  has  a  weight  of  50  Ib.  placed 
upon  it  so  that  it  causes  an  increase  in  pressure  on  the  support  at  one  end 
of  20  Ib.     It  is  also  found  that  the  same  pressure  is  produced  at  this  end 
by  a  weight  of  60  Ib.  placed  3  ft.  farther  from  this  end.     How  long  is  the 
beam?  Ant.  45  ft. 

Suggestion.     Consider  the  support  where  the  increase  is  20  Ib.  as  the 
fulcrum.     Then  there  is  a  pressure  of  30  Ib.  at  the  other  end. 

Let         y  =  the  length  of  the  beam 
and  x  =the  distance  the  50-lb.  weight  is  from  the  fulcrum. 

Then  30y  «  50x. 

Similarly,  when  the  60-lb.  weight  is  applied, 
40?/=60(z+3). 

37.  A  glass  full  of  water  weighs  18  ounces.     When  the  same  glass  is 
full  of  sulphuric  acid  of  specific  gravity  1.75  it  weighs  27  ounces.     Find 
the  weight  of  the  glass  when  empty.  Ana.  6  ounces. 

38.  Given  two  grades  of  zinc  ore,  the  first  containing  45%  of  zinc,  and 
the  second  25%  of  zinc.     Find  how  many  pounds  of  each  must  be  taken 
to  make  a  mixture  of  2000  Ib.  containing  40%  of  zinc. 

Ans.  1500  Ib.  of  45%  ore  and  500  Ib.  of  25%  ore. 

Suggestion.      Let  x=  number  of  pounds  of  45%  ore, 
and  y=number   of  pounds  of  25%  ore. 

Then  z+y  =  2000, 

and         0. 45* +0.25y  =  0.40X2000. 

39.  The  sum  of  the  three  sides  of  a  triangle  is  65  in.     If  the  second  side 
is  5  in.  longer  than  the  first  and  7  in.  shorter  than  the  third,  find  the 
length  of  each  side.  Ans.  16  in.;  21  in. ;  28  in. 


EXPONENTS,  POWERS,  AND  ROOTS 

262.  General  statement. — In  previous  chapters,  we  have 
used  positive  integral  exponents.     In   Art.   185,  a  positive 
integral  exponent  was  denned  as  showing  how  many  times  the 
base  is  to  be  used  as  a  factor. 

The  negative  exponent,  the  fractional  exponent,  and  the 
zero  exponent  are  other  kinds  of  exponents  that  occur  in 
mathematics.  They  will  be  dealt  with  to  some  extent  in  the 
present  chapter.  Logarithms  are  exponents  that  will  be 
considered  in  Chapter  XXXIV. 

The  few  facts  and  theorems  concerning  exponents,  given  in 
this  chapter,  are  not  intended  to  be  a  complete  treatment  of 
the  subject,  but  they  are  sufficient  for  what  follows,  and  will 
give  the  proper  viewpoint  for  logarithms. 

It  should  be  carefully  noted  that  the  definition  of  a  positive 
integral  exponent  cannot  apply  to  an  exponent  that  is  negative, 
zero,  or  a  fraction.  For  instance,  in  82,  the  2  shows  that  8 
is  taken  twice  as  a  factor;  but  in  8*,  the  ^  can  in  no  sense  show 
how  many  times  8  is  taken  as  a  factor.  To  say  that  8  is 
taken  |  times  as  a  factor  is  to  make  a  meaningless  statement. 

We  shall  therefore  find  it  necessary  to  enlarge  the  definition 
of  an  exponent  so  as  to  make  it  include  these  new  kinds  of 
exponents. 

263.  Laws  of  exponents. — The  law  of  exponents  in  mul- 
tiplication  (see  Art.   216)   has  been  stated  and  proved    for 
positive  integral  exponents.     It  may  be  restated  here  in  symbols 
as  follows : 

am-an  =  am+n. 

Likewise  the  law  of  exponents  in  division  (see  Art.  223)  has 
been  stated  for  positive  integral  exponents.  A  restatement  in 
symbols  is  as  follows: 

am-^-an  =  am"n, 

where  m  is  greater  than  n.     When  ra  =  n  it  is 

an  +  an  =  l. 
327 


328  PRACTICAL  M AT HEMATICS 

In  finding  the  power  of  a  power  the  exponents  are  multiplied. 
That  is,  (am)n  =  omn.  This  is  found  from  the  definition  of  a 
positive  integral  exponent. 

Thus,  (am)n=amamam-  •  •  to  n  factors  =am+m "•"""•  •  •  to  n  terms  «a*". 
A  numerical  illustration  is  (a*)s=a*-a'-a'=»o18. 

The  power  of  a  product  is  the  same  as  the  product  of  the  powers 
of  the  factors. 

That  is,  (abc-  •  -)n=anbncn-  •  •• 

A  numerical  illustration  is  (2-34)3  =  2»-33-43. 

The  power  of  a  fraction  equals  the  power  of  the  numerator 
divided  by  the  power  of  the  denominator. 

fa\  m     a* 


That  is,    ^     -~ 

2\  *    2*     8 


A  numerical  illustration  is  •„•    —  „, 
\o/        o 

//  we  take  the  root  of  a  power,  we  have  the  inverse  of  the  opera- 
tion by  which  we  get  amn,  and  have  •^/am  =  am^n. 
A  numerical  illustration  is  -^/3*=344-2  =  32  =  9. 

The  definition  of  a  positive  integral  exponent  gives  us  the 
right  to  state  the  above  only  when  m-r-n  is  an  integer. 
A  summary  of  the  six  laws  of  exponents  mentioned  is: 

(1)  aman  =  am+n. 

(2)  am+an  =  am-n. 

(3)  (am)n  =  aron. 

(4)  (a-b-c  -  •  -  )m  =  ambmcm  -  -  -. 


(G) 

Example  1.     Find  the  fourth  power  of  3azx3y4. 


Example  2.     Find  the  cube  of 


2t/2 
33(a5)3(b4)3x*  =  27a9bl*x* 

2%2)3  8t/« 

Example  3.     Find  the  cube  root  of  36&V2. 

;«+ 3=3263a;4. 


EXPONENTS,  POWERS,  AND  ROOTS  329 

EXERCISES  92 

Raise  to  the  indicated  powers,  or  find  the  root  indicated. 

1.  (oV)4.  Ans.  a8?/12. 

2.  (2?/3x4)6.  Ans.  64?/18x24. 

3.  (3a26)3.  Ans.  27a663. 

4.  (-4a2x6)3.  Ans.  -64a6x18. 
6.  (-3a3b2x)4.  Ans.  81a1268x4. 

6.  (— x4?/6)2.  Ans.  x8^12. 

7.  (3°6C)3.  Ans.  S3^30. 

8.  (4acx2)6.  Ans.  46a6cx26. 

27x3 
Ans. 

Ans. 


12569 
Ans'  256c'«x8' 


Ans. 
Ans.  7x« 


4b3 


264.  Zero   exponent.  —  If   we    assume   that   am  -;-  an  =  am~n 
holds  when  m  =  n,  we  have  aTO  -r-  an  =  am~n  =  a°,  for  w  —  n  =  0 

Also  a"1  -s-  a"  =  1  by  Art.  223,  if  m  =  n. 

:.  a°=i. 

Making  the  above  assumption  is  the  same  as  stating  by 
definition:  Any  number  other  than  zero  affected  by  a  zero  ex- 
ponent equals  one. 

Examples.     (1)    0.8°  =  1. 

(2)  100°  =  1. 

(3)  0.01°=  1. 

265.  Negative  exponent.  —  Assuming  that  the  law  am+an  = 
am-n  holds  when  n  is  greater  than  m,  we  have 


330  PRACTICAL  MATHEMATICS 


a" 

Similarly  a~"  =  — 
on 

By  definition  then,  a  number  affected  by  a  negative  exponent 
equals  1  divided  by  the  same  number  affected  by  a  positive  ex- 
ponent, equal  in  absolute  value  to  the  negative  exponent. 

Examples.     (1)  2~3  =  o3  =  o' 

(2)  4->  =  —  =  — 
43     64- 

266.  Fractional  exponent.  —  If  we  apply  the  law  \s'am  =  am  +  n} 
when  m  and  n  have  any  values,  we  have 


an.     Also 

By  definition  then,  a  fractional  exponent  indicates  a  root. 
The  denominator  is  the  index  of  the  root,  and  the  numerator  is 
the  exponent  of  a  power. 

m 

A  form  like  an  means  either  -\/amor  (v^a)m-  That  is,  the 
number  a  may  be  raised  to  the  rath  power  and  then  the  nth 
root  taken,  or  the  nth  root  may  be  taken  first  and  then  the 
result  raised  to  the  mth  power. 

Thus,  8'  =  ^/8~'  =  ^64  =  4,  or  8J 
Examples.     (1)    4§  =\/4  =  2. 

(2)  64*  =  -^64  =4. 

(3)  32* 


EXERCISES    93 

Find  the  values  of  the  following: 

1.  10°.                                   Ans.  1.  6.   16-*.  Ans.  \. 

2.  4-'                                  Ans.  A.  7.  27>.  Ans.  81. 

3.  10'*.                   ATM.  roooJinnnj.  8.  512~J.  Ans.  &. 

4.  8-'.                               Ans.  ,},.  9.  dh)-'.  Ans.  25. 

5.  16*.                                  Am.  2.  10.   100*.  Ans.  1000. 


EXPONENTS,  POWERS,  AND  ROOTS  331 

11.  1000-i.  Ans.  xta.     14.  (4°)«.  Ans.  1. 

12.  10°-32.  Ans.  9.     16.  64~5  Ans.  T1fei. 

13.  (3°)-3.  Ans.  1.     16.  9*.  Ans.  243. 

17.  Divide  a~*x~*  by  cr3.  Ans.  ax~\ 

18.  Multiply    a~»   by  a8.  Ans.  a*X 

19.  Multiply  3a*  &*  by  4a*  6*.  Ans.  12a«  b*. 

267.  Exponents  used  in  writing  numbers. — Often  in  engi- 
neering subjects  we  see  such  expressions  as  4.25  X109,  or 
726X10~8.  These  forms  are  the  result  of  an  attempt  to  write 
certain  expressions  in  a  shortened  form.  Keeping  in  mind 
the  meaning  of  an  exponent, 

4.25  X 109  =  4.25  X  1,000,000,000  =  4,250,000,000. 

This  last  number  means  the  same  as  the  given  expression, 
but  in  some  ways  is  not  so  convenient  a  form  to  handle, 
neither  is  it  so  easily  compared  with  others  of  its  kind  as  is 
the  first  form.  Neither  does  the  eye  catch  its  value  as  quickly 
as  it  does  that  of  the  shortened  form. 

The  form  726  X  10~8  =  726  X  ^  =  726  X  i  <nr oi  OTTDT  = 

0.00000726. 

EXERCISES  94 

Express  the  following  in  the  common  notation. 

1.  3.5  X107.  Ans.  35,000,000. 

2.  8.67  X1011.  Ans.  867,000,000,000. 

3.  523 X10-8.  Ans.  T<juf!iihnn7=  0.00000523. 

4.  4.786 X10-".  Ans.  TOO utrVo8o6(jooTO  =  0.000000004786. 

5.  9.376  X 10-6.  Ans.  Tru?!J!hnnr= 0.000009376. 

6.  4.673  X108.  Ans.  467,300,000. 

7.  4.37  X1012.  Ans.  4,370,000,000,000. 
Express  the  following  in  the  shortened  notation : 

8.  4,768,000,000.  Ans.  4.768  X109. 

9.  23,600,000,000.  Ans.  2.36  X1010. 

10.  37,600,000,000.  Ans.  37.6  X109. 

11.  i^oruo.  Ans.  3.67 X 10-". 

12.  TcrasWo.  Ans.  4X10~7. 
13.0.000000367.  Ans.  3.67  X1Q-7. 

14.  0.000000004676.  Ans.  4.676  XlO"9. 
16-  i&fflv  Ans.  2.798  X10~«. 


CHAPTER  XXXI 
QUADRATIC  EQUATIONS 

268.  Definitions. — An  equation  that  contains  the  square  of 
the  unknown  number  and  no  higher  power  of  it  is  a  quadratic 
equation. 

A  pure  quadratic  equation  is  one  that  has  the  square  only 
of  the  unknown  number,  as  2x2  =  4. 

An  affected  quadratic  equation  is  one  that  has  both  the 
square  and  the  first  power  of  the  unknown,  as  x2+3x  =  10. 

THE  PURE  QUADRATIC  EQUATION 

269.  Solution. — The  solution  of  the  pure  quadratic  equation 
is  the  same  as  that  of  the  simple  equation  until  the  value 
of  the  square  of  the  unknown  is  found.     The  next  step  is  to 
take  the  square  root  of  each  member  of  the  equation. 

Example  1.     Solve  3x2+8  =  7x2  -  8  for  x. 
Solution. 

(1)  Given  equation,         3x2+8  =  7z2-8. 

(2)  Transposing,          3z2-7z2= -8-8. 

(3)  Collecting  terms,         -4x2=-16. 

(4)  Dividing  by  —4,  #2  =  4. 

(5)  Extracting  the  square  root  of  both  x2  and  4,     x=  ±2. 
Here  we  use  the  sign  ±,  and  it  indicates  that  the  answer 

i&  either  a  +2  or  a  -2.     (See  Art.  237.) 

Since  the  product  of  two  minus  numbers  gives  the  same  value 
as  the  product  of  the  same  plus  numbers,  we  may  always  call 
a  square  root  either  plus  or  minus. 

Thus,  \/4=  ±2,  Vb*=  ±b. 

Example  2.     Find  the  radius  of  a  circle  whose  area  is 4392  in.2 

Solution.     (1)  Let  r  stand  for  the  radius, 

(2)  then  3. 1416r2  =  area  of  circle. 

(3)  But  4932  in.2  =  area  of  circle. 

(4)  .'.  3.1416r2  =  4932,  by  axiom  (5). 

(5)  r2=  1569.9007. 

(6)  r=  ±39.622. 
332 


QUADRATIC  EQUATIONS  333 

It  should  be  noticed  that  the  radius  cannot  be  negative, 
hence  the  only  value  of  r  permissible  is  +39.622. 

EXERCISES  96 

Solve  for  x  in  exercises  1  to  10  and  test. 

1.  7z2+25=2z2  +  150.  Ans.  ±5. 

2.  3x2-10-x2  =  12+4a-2-54.  Ans.  +4. 

3.  53-7z"+27  =  -2.c2.  Ans.  ±4. 

Ans.   ±2. 
Ans.  ±3. 


. 

15  o 

,    2x      5      7x     21 

6.  -_-  -  -:-  =  -jr-  -  -r-  Ans.  ±6. 

3       4x      9       4x 


7.  2x(5-x)+8x2  =  10(x+2).  Ans.  ±1.826-. 

^.±1.826-. 


9.  (3x+6)(3x-6)  =  (2x+5)(2z-5).  Ans.  ±1.483+. 

Ans.  ±2. 
x—l  x+1 

5  58 

11.  Given \-  -. =  «,  find  the  value  of  y.  Ans.   ±  1. 

4+t/     4-y     3' 

Wv2  iFor 

12.  Given  F  = .  find  the  values  of  v.  Ans.    ±\hn^' 

gr  '  \  W 

13.  (a)  Given  S  =  $gt*,  find  values  of  t.     (b)  If  5  =  5280  and  fir  =  32.2, 

I2S 
find  values  of  t.  Ans.  (a)  ±  \[— J  (&)  + 18.11. 

14.  Given  I——,  solve  for  t.  Ans.   +  \  —  • 

T  \  ^ 

mna      ,       ..       ,  /mna 

16.  Given  F=—^-,  solve  for  a.  Ans.    ± 


16.  Given  n2  =  /2  ,  ,  solve  for  <S,  /,  and  n. 


, KSH  KSH 

Ans.  ±  \—f^r ',  ±  \"^~T  '»  ±  \~l2Z" 

17.  Find  the  radius  of  a  cylinder  of  altitude  12  ft.  and  volume  1400 
cu.  ft.      (Use  formula  V  =  hirr2,  substitute  values  and  solve  for  r.) 

Ans.  6.09+  ft. 

18.  Find  the  diameter  of  a  right  circular  cone  of  altitude  20  in.  and 
volume  145  in.3  Ans.  5.26+  in. 


334  PRACTICAL  MATHEMATICS 

THE  AFFECTED  QUADRATIC  EQUATION 

270.  Solution  by  factoring. — An  affected  quadratic  equation 
in  x,  when  simplified,  can  have  a  term  in  x2,  a  term  in  x,  a 
term  not  containing  x,  and  no  other  term.     Thus,  x2  —  5z  =  —  0 
is  such  an  equation. 

The  solution  of  this  equation  has  been  discussed  in  Art.  237. 
The  steps  in  the  solution  of  a  quadratic  equation  that  can  be 
solved  by  factoring  are  restated  here. 

(1)  Simplify  the  equation. 

(2)  Transpose  all  terms  to  the  first  member  of  the  equation. 

(3)  Factor  the  expression  in  the  first  member. 

(4)  Equate  each  factor  to  zero. 

(5)  Solve  each  of  these  equations. 
Example.     Solve  x2+23x  =  - 102. 

Solution.     (1)  Given  equation,  x2+23x= —102. 

(2)  Transposing,  x2 + 23x  + 102  =  0. 

(3)  Factoring,  (x+6)(x+17)  =0. 

(4)  .'.  x+6  =  0orx+17  =  0. 

(5)  .'.  x=-6orz=-17. 

EXERCISES  96 

Solve  and  test. 

1.  zl  +  15z+56  =  0.  Ans.  -7  or  -8. 

2.  x1- 17a;+72=0.  Ana.  8  or  9. 

3.  z2-16z=36.  Ans.  -2  or  18. 

4.  5x-6=x2.  Ans.  2  or  3. 

5.  z2+2z=24.  Ans.  4  or  -6. 

6.  *2+26z  +  160=0.  Ans.  -10  or  -16. 

7.  z2+3z  =  54.  Ans.  -9  or  6. 

271.  It  was  stated  in  Art.  235  that  there  are  many  trino- 
mials we  had  not  learned  to  factor.     It  is  necessary  then  to 
have  some  other  method  than  factoring  for  solving  affected 
quadratic  equations. 

272.  Completing   the    square. — In    Arts.  229  and  230  we 
learned    the    form    of   a  trinomial    square.    The   first   and 
last  terms  are  perfect  squares  of  monomials,  and  the  middle 
term  is  twice  the  product  of  the  square  roots  of  the  first  and 
last  terms.     Keeping  this  in  mind  then,  we  can  find  the  last 
term  if  we  know  the  first  two- 


QUADRATIC  EQUATIONS  335 

Thus,  if  we  know  that  a2+2a6  are  the  first  two  terms,  we  can 
find  the  third  term  by  taking  the  square  of  the  quotient  ob- 
tained by  dividing  the  second  term  by  twice  the  square  root  of 
the  first.  Twice  the  square  root  of  the  first  term  is  2a. 
The  second  term  divided  by  this  gives  6;  which  squared  is  62, 
the  third  term.  We  say  then  that  we  have  completed  the 
square,  which  is  a2+2«6+62. 

Example.     Complete  the  square  of  x2+4x. 

Twice  the  square  root  of  x2  is  2x,  4x-^-2x  =  2,  and  the  square 
of  2  is  4.  Hence  the  completed  square  is  xz+4x+4. 

When  the  coefficient  of  a;2  is  1,  all  that  is  necessary  is  to  add 
the  square  of  one-half  the  coefficient  of  x. 

Complete  the  square  of  each  of  the  following: 

1.  x-+2x.  Result  z2+2:r+l. 

2.  x2-2x.  Result  x2  -2x+l. 

3.  z2-3z.  Result  z2-3z+f. 

4.  x2-Wx.  Result  z2-10z+  25. 

5.  a2-10a&.  Result  a2-  10a&+25&2. 

273.  Solution  by  completing  the  square.  —  We  can  best 
show  how  this  is  applied  to  the  solution  of  an  equation  by  an 
example. 

Example     1.  Given  xz-^-4x  =  12}  find  x. 

(1)  Completing  the  square  of  the  first  member  we  have 
z2+4z+4.  Since  4  is  added  to  x2+4x,  we  must  also  add  4  to 
12  in  order  that  the  equality  may  be  true. 


(2)  Extracting  the  square  root  of  both  members,  we  have 

z+2=±4. 

(3)  Transposing,  x=  —  2+4. 

(4)  :.x  =  2  or  -6. 

Each  of  these  results  checks  the  original  equation  and  is 
therefore  a  root  of  the  equation. 

Example    2.  Given  x2-\-12x=  —35,  find  x. 
Solution.     (1)  Given  equation,  x2+12x=  —  35. 

(2)  Completing  square,      £2+  12x4-36  =  1. 

(3)  Extracting  roots,  x-\-Q=  +1. 

(4)  :.x=  —7  or  —5. 


336  PRACTICAL  MATHEMATICS 

Example  3.     Given  2r2+4r  =  48,  find  r. 
Solution.     (1)  Given  equation,  2r2+4r  =  48. 

(2)  Dividing  by  2,  r*+2r  =  24. 

(3)  Completing  square,  rl+2r+l  =  25. 

(4)  Extracting  roots,  r+l=±5. 

(5)  .'.r  =  4  or  -6. 
Example  4.     Given  3z2  — 7z  =  6,  find  x. 
Solution.     (1)  Given  equation,  3z2  —  7z  =  6. 

(2)  Dividing  by  3,  z2-fz  =  2. 

(3)  Completing  square,          x*-lx+ j|  =W- 

(4)  Extracting  roots,  x  —  |=±Y- 

(5)  /.z=3or-i 

It  should  be  noticed  that  we  always  divide  by  the  coefficient 
of  x2,  unless  it  is  1,  before  completing  the  square. 

Example  5.     Given  9x  =  4  —  3z2,  find  x. 

Solution.     (1)  Given  equation,  9z  =  4  —  3x2. 

(2)  Transposing,  3z2+9z  =  4. 

(3)  Dividing  by  3,  z2+3z  =  ^. 

(4)  Completing  square,  z2+3z+f  =  s  +  f  =  H- 

(5)  Extracting  roots,  x+f=±Vft=± 1-893  ~  • 

(6)  :.x  =-|±1.893-. 

(7)  x=  0.393  or  -3.393. 

The   method   of  solving   an   affected   quadratic    equation 
may  be  summarized  in  the  following: 

RULE.     (1)  Reduce  the  equation  to  the  form  x2+bx  =  c. 

(2)  Complete  the  square  by  adding  to  both  members  the  square 
of  one-half  the  coefficient  of  x. 

(3)  Equate  the  square  root  of  the  first  member  to  ±  the  square 
root  of  the  second  member,  and  solve  the  two  equations  thus  formed. 

EXERCISES  97 

Solve  for  x  by  completing  the  square. 

1.  *2-4z=60.  An*.  10  or  -6. 

2.  z«  +  llz=-24.  Ans.  -8  or  -3. 

3.  3j2-10x+3=0.  Ans.  3  or  |. 

4.  3()x*  -36x  -7=0.  Ans.  i  or   -  J. 
6.   10z'+7z  =  12.  Ans.  i  or    -?. 
6.  j»+4z=2.                                                  Ans.  0.4494   or    -4.4494. 


QUADRATIC  EQUATIONS  337 

7.  z2+2z=2.  Ans.  0.732  or   -2.732. 

8.  4z2-4z  =  7.  Ans.  1.9142  or    -0.9142. 

9.  9z2+6z-17  =  0.  Ans.  1.0809  or    -1.7475. 
10.  3x2  +  121=44x.  Ans.  11  or  V- 


274.  Solution  of  the  affected  quadratic  equation  by  the 
formula.  —  If  we  solve  the  general  quadratic  equation  oo;2+ 
bx+c  =  Q,  we  shall  have  a  formula  which  may  be  used  to  find 
the  value  of  the  unknown  in  any  quadratic  equation. 

Example  1.     Given  axz+bx-{-c  =  Q,  find  x. 

Solution. 

(1)  Given  equation,  ax-  +  bx  +  c  =  Q. 

(2)  Transposing,  az2  +  &z  =—c. 

b  c 

(3)  Dividing  by  a,  x*-\  —  x=  --- 

b        b2       b2      c 

(4)  Completing  square,  x2+-x  +  -.  -^  =  -r^>  --- 

a       4a2     4a2     a 


.    b  I 

(o)  Extracting  roots,  £+K-=+A/ 

b   ,      Ib2-4ac     -b± 


b2  —  4ac 


This  last  form  is  a  formula  that  can  be  used  in  solving  any 
quadratic  equation. 

In  using  the  formula,  care  must  be  taken  as  regards  the 
signs.  They  must  be  considered  a  part  of  the  values  of 
a,  b,  and  c. 

Thus,  in  3z2-2z-4  =  0,  o  =  3,  b=  -2,  and  c=  -4. 

Example  2.  Substitute  in  the  formula  of  example  1,  and 
find  the  value  of  x  in  6z2+17o;+7  =  0. 

Solution,     Here  a  =  6,  b  =  17,  and  c  =  7. 

Substituting  these  values  in  the  formula, 


-_  _  _j  , 

^6~  ~T2~ 

Example  3.     By  the  formula,  find  the  value  of  x  in  6z2-|- 
8z-30=0. 

Solution.     Here  a  =  6,  6  =  8,  and  c=  —  30. 


•      _-8  +  V82-4-6(-30)=-8±28 


338                          PRACTICAL  MATHEMATICS 

EXERCISES  98 
Solve  the  following  equations  by  any  method. 

1.  zl  -3x  -  -2.  Ans.  1  or  2. 

2.  z»+5x=-6.  Ans.  -2   or    -3. 

3.  z'-ar-O.  4n*.  3  or  -2. 

4.  t/*-2j/  =  168.  An*.  14  or   -12. 

5.  yl+2j/  =  120.  Ana.  10  or    -12. 

6.  j/2-22j/=48.  Ans.  24   or    -2. 

7.  5r2  +  12r  =  9.  ATM.  |  or  -3. 

8.  2r2+5r=-2.  4ns.  -J   or    -2. 

9.  7a2+9ct  =  10.  ATM.  ?  or  -2. 

10.  7x*+2x  =  32.  An*.  2  or  -2?. 

11.  F+i-;j  =0.  Ans.  2  or  I 

V        £ 

12'?+f=T  Ans.  71  or  -WL 

96*  6            96 

13.  x*+4bx=-j--  Ans.  -  or  — ^' 


14.  Given  S  =  Vt  +  $(7<2,  solve  for  V,  g,  and  <. 

2S-gt\    2S-2VI.     -V±VV*+2gg. 

2t      '          <2       '  <; 

15.  Given  !T  =  27rr/i+2irr2,  solve  for  r.         Ans.  - 


16.  Find  three  consecutive  numbers,  such  that  their  sum  is  one-third 
of  the  product  of  the  first  two.  Ans.  9,  10,  11  or  —  1,  0,  1. 

17.  Find  two  numbers,  one  of  which  is  5  times  the  other,  and  whose 
product  is  4500.  Ans.  150  and  30,  or  -150  and  -30. 

18.  A  walk  containing  784  ft.2  is  to  be  built  around  a  garden  50  by  49 
ft.     How  wide  must  the  walk  be?  Ans.  3  ft.  8  J  in.  nearly. 

19.  There  are  as  many  square  feet  in  the  surface  of  a  certain  sphere  as 
there  are  cubic  feet  in  its  volume.     Find  its  radius.  Ans.  3  ft. 

Solution.     Let  r  =  number  of  feet  in  radius. 
Then  4nr*  =  number  of  square  feet  in  surface, 
and  3irr3  =  number  of  cubic  feet  in  volume. 


4rr8-12irrJ=0. 

Factoring,  4nrs(r-3)  =0. 

.•.4irr2=0  and  r-3=0, 

or  r  =  0  and  r  =  3. 

20.  Same  as  last  exercise,  but  substitute  cube  for  sphere  and  find  the 
edge.  Ans.  6   ft. 

21.  Find  two  numbers  whose  sum  is  25  and  whose  product  is  144. 

Ana.  9  and    16. 

22.  Divide  71  into  two  parte,  the  sum  of  the  squares  of  which  is  2561. 

Ans.  40  and  3L 


QUADRATIC  EQUATIONS  339 

23.  If  a  train  traveled  5  miles  an  hour  faster,  it  would  take  2  hours 
less  to  go  420  miles.     Find  the  rate  of  the  train. 

Ans.  30  miles  an  hour. 

24.  What  is  the  radius  of  a  circle  whose  area  is  doubled  when  the 
radius  is  increased  by  2  ft.?  Ans.  4.828  ft. 

25.  What  is  the  diameter  of  a  circle  whose  area  is  multiplied  by  3 
when  the  diameter  is  increased  by  2  ft.?  Ans.  2.732  ft. 

26.  The  height  of  a  right  circular  cylinder  is  6  in.,  and  the  entire  area 
is  100  in.2     Find  the  radius  of  the  base.  Ans.  1.99+  in. 

Suggestion.     Let  r  stand  for  radius  of  base. 
Then  2irr'2  +  l2wr  is  the  entire  area. 


100 

or  r2+6r  =  --  =  15.915. 

2ir 

27.  The  differences  between  the  hypotenuse  and  the  two  sides  of  a 
right  triangle  are  3  and  6  ft.  respectively.     Find  the  sides  and  the  area 
of  the  triangle.  Ans.  15  ft,,  12  ft.,  9  ft.,  and  54  sq.  ft. 

28.  The  hypotenuse  of  a  right  triangle  is  4  in.  longer  than  one  leg,  and 
8  in.  longer  than  the  other.     Find  the  lengths  of  the  three  sides. 

Ans.  20  in.;  16  in.;  12  in. 

29.  The  circumference  of  the  hind  wheel  of  a  wagon  is  5  ft.  more  than 
that  of  the  front  wheel.     If  the  hind  wheel  makes  150  fewer  revolutions 
than  the  front  wheel  in  going  one  mile,  find  the  circumference  of  each 
wheel.  Ans.  16  ft.  and  11  ft. 

30.  An  aeroplane  which  is  at  an  altitude  of  1200  ft.  and  moving  at  the 
rate  of  100  miles  per  hour  in  a  northerly  direction  drops  a  bomb.     Dis- 
regarding the  resistance  of  the  air,  where  will  the  bomb  strike  the  ground? 

Ans.  1270  ft.  north  of  starting  point. 

Suggestion.  Find  the  number  of  seconds  it  will  take  the  bomb  to  reach 
the  ground  from  the  formula  s  =  %gt~,  where  s=height  of  aeroplane  in 
feet,  g  =32,  and  t  =time  in  seconds.  This  gives  t  =  8.66  seconds. 

Then  the  bomb  will  strike  as  many  feet  north  of  the  starting  point  as 
the  aeroplane  will  travel  in  8.66  seconds. 


CHAPTER  XXXII 
VARIATION 

276.  General  statement. — We  depend  upon  the  ideas  dealt 
with  in  variation  for  most  of  our  physical  laws  and  formulas. 
The  meaning  of  many  of  the  formulas  already  used  will  be 
made  much  clearer  by  considering  their  meaning  from  the 
viewpoint  of  variation. 

The  relations  considered  in  variation  are  those  considered 
in  ratio  and  proportion.  In  many  ways,  however,  it  will  be 
found  that  the  methods  used  in  variation  are  more  convenient 
than  the  methods  of  ratio  and  proportion.  Familiarity  with 
them  gives  the  student  another  powerful  mathematical 
instrument. 

For  the  principles  of  ratio  and  proportion  used  in  the  present 
chapter  the  student  is  referred  to  Chapter  VIL 

276.  Constants   and   variables. — A   number   whose   value 
does  not  change  is  called  a  constant. 

In  mathematical  problems,  certain  constants  occur  that  are 
always  the  same. 

The  value  of  *,  the  ratio  of  the  circumference  to  the  diameter  of  a  circle, 
is  such  a  constant. 

There  also  occur  other  constants  that  do  not  change  in  the 
same  problem,  but  which  may  have  another  value  in  a  dif- 
ferent problem. 

The  radius  of  a  circle  and  the  side  of  a  square  are  such  constants. 

A  variable  is  a  number  that  may  take  an  unlimited  number 
of  values. 

For  example,  the  number  expressing  a  person's  age,  the  height  of  a 
growing  corn  stalk,  or  the  distance  a  moving  train  is  from  a  station  it  has 
just  left. 

277.  Direct  variation. — The  idea  of  variation  is  a  very  com- 
mon one.     Nearly  everything  is  affected  by  its  surroundings; 
that  is,  things  vary  according  as  something  else  varies. 

340 


VARIATION  341 

The  growth  of  a  tree  depends  upon  the  amount  of  light  it 
receives;  the  more  light  it  receives,  the  faster  it  grows,  if  other 
conditions  are  favorable.  In  such  a  case  we  say  its  growth 
varies  directly  as  the  amount  of  light. 

The  amount  of  pay  a  man  gets  varies  directly  as  the  time 
he  works;  that  is,  the  longer  he  works  the  more  pay  he  receives. 

Definition.  If  two  numbers  are  so  related  that  their  ratio 
is  constant,  that  is,  if  either  increases  the  other  increases,  or 
if  either  decreases  the  other  decreases,  the  two  numbers  are 
said  to  vary  directly  as  each  other. 

278.  Mathematical  statement.  —  Just  as  with  many  other 
ideas,  there  is  a  mathematical  way  of  expressing  the  idea  of 
variation.  When  the  ideas  are  so  expressed,  they  can  be 
combined  and  handled  according  to  the  rules  of  mathematics. 
In  this  manner,  new  relations  are  seen  and  new  results  obtained. 

The  sign  <*  means  "varies  as." 

If  x  and  y  are  two  variables  that  vary  directly  as  each  other, 
it  may  be  written  in  the  shorthand  form  x  oc  y.  It  may  also 
be  written  in  the  form  x  =  ky.  This  last  is  an  equation  stating 
that  the  ratio  of  x  to  y  is  a  constant  k.  The  equation  is  the 
form  most  often  used. 

Example  1.  If  a  train  is  moving  away  from  a  station  at  a 
uniform  rate,  express  the  relation  between  its  distance  d 
from  the  station  and  the  time  t  since  it  left  the  station. 

Here  evidently  the  distance  d  varies  directly  as  the  time  t. 


The  student  should  consider  very  carefully  the  meaning  of 
the  constant  k  in  the  different  examples  taken  up.  Here 
the  k  evidently  represents  the  uniform  rate  per  unit  of  time. 
If  the  time  is  in  minutes,  k  stands  for  the  rate  per  minute. 
This  rate  may  be  given  in  feet,  rods,  miles,  or  any  other  unit 
of  length.  The  k  then  depends  for  its  value  upon  the  kind 
of  units  used,  but  this  is  not  the  only  thing  that  may  change 
the  value  of  k. 

Example  2.  If  two  numbers  x  and  y  vary  directly  as  each 
other,  and  when  x  =  10,  ?/  =  4,  find  x  when  i/  =  25. 

Solution.  Since  the  relation  between  x  and  y  is  direct, 
x  =  ky. 


342  PRACTICAL  MATHEMATICS 

Substitute  the  values  of  x  and  y  given,  and 


.'.  x  =  2\y  is  the  relation  between  the  variables,  and  if 
?/  =  25  we  have  z  =  2^-25  =  62$.  Ans. 

Such  a  relation  as  x  =  2$y  is  often  spoken  of  as  a  law. 

Example  3.  The  space  passed  through  by  a  body  falling 
freely  from  a  distance  above  the  ground  varies  as  the  square 
of  the  time.  If  s  =  the  space  in  feet  and  <  =  time  in  seconds, 
write  the  law.  Find  the  value  of  k,  if  the  body  will  fall  402.5 
feet  in  5  seconds. 

Solution.  Since  the  variation  is  directly  as  the  square  of 
the  time,  s  =  kt2.  Substituting  values  of  s  and  /, 

402.5  =  k  -52. 
.'.  fc  =  16.1. 

Hence  the  law  or  formula  to  find  the  distance  in  feet  a  body 
will  fall  in  t  seconds  is  s  =  16.  K2.  This  is  usually  written 
s  =  $gt2,  where  0  =  32.2. 

279.  Inverse  variation.  —  Consider  a  horizontal  beam,  resting 
at  each  end  on  a  support,  and  having  a  weight  at  its  midpoint. 
The  size  of  the  weight  it  will  support  depends  upon  its  length  ; 
but  here  the  longer  the  beam,  the  less  it  will  support.     In  such 
a  case  the  variation  is  said  to  be  indirect  or  inverse. 

The  resistance  to  an  electric  current  is  less  in  a  large  wire 
than  in  a  small  one  of  the  same  material;  the  resistance  varies 
inversely  as  the  size  of  the  wire  in  cross  section. 

The  intensity  of  the  light  from  a  lamp  decreases  as  we  go 
away  from  it.  Here  the  variation  is  an  inverse  one,  but  the 
intensity  of  illumination  is  not  one-half  as  much  when  the 
distance  is  doubled.  It  decreases  inversely  as  the  square  of 
the  distance;  that  is,  the  intensity  is  one-fourth  as  much  at 
twice  the  distance. 

Definition.  One  number  varies  inversely  as  another  when 
their  product  is  constant.  That  is,  if  either  increases,  the 
other  decreases,  or  if  either  decreases,  the  other  increases. 

280.  Mathematical    statement.  —  The    shorthand    way    of 

writing  the  fact  that  x  varies  inversely  as  y  is  x  «  — 

J 

The  equation  form  is  x  =  ->  or  xy  =  k. 

y 


VARIATION  343 

Example  1.  If  x  varies  inversely  as  y,  state  the  law  and  find 
the  value  of  the  constant  if  x  =  10  when  y  =  \  . 

Solution.  xy  =  k  is  the  mathematical  statement  of  the 
variation. 

Substituting,  lQ-^  =  k.     .'  .  k  =  5. 

If  this  value  of  k  is  used,  the  law,  or  equation,  becomes 


Example  2.     If  Z  =  TITO-J  find  y  from  the  law  of  example  1. 
Substituting  in  the  law,  T-^y  =  5. 


.'.  ?/  =  500.  Ans. 

281.  Joint  variation.  —  Definition.  One  number  varies 
jointly  as  two  or  more  other  numbers  when  it  varies  directly 
as  the  product  of  the  others.  Thus,  x  varies  jointly  as  u 
and  v  when  x  =  kuv. 

A  number  may  vary  directly  as  one  number  and  inversely 
as  another.  It  then  varies  as  the  quotient  of  the  first  divided 
by  the  second. 

V 

Thus,  if  x  varies  directly  as  y  and  inversely  as  z,  it  is  written  x  —  k-  • 

2 

EXERCISES  99 

1.  If  x  oc  y  and  when  a;  =  20,  y  =  QO,  write  the  equation    between  x 
and  y.  Ans.  3x  =  y. 

2.  If  b  oc  d  and  when  6  =  10,  d  =  15,  find  b  when  d  =80.         Ans.  53  \. 

3.  If  x  varies  jointly  as  y  and  z,  and  when  y  =  G  and  0  =  2,  x  =  120, 
find  the  constant.     Find  y  when  z  =  200  and  3  =  15. 

Ans.  fc  =  10;  y  =  \\. 

4.  The  area  A  of  a  triangle  varies  jointly  as  the  base  &  and  the  altitude 
a.     Write  the  law  if  when  a  =  6  in.  and  6  =  4  in.,  A  =12  in.2     What  will 
be  the  area  when  the  base  is  25  in.  and  the  altitude  is  40  in.  ? 

Ans.  A=%ab;  500  in.2 
Remark.     The  law  here  is  the  well-known  formula  for  the  area  of  a 

triangle,  but  we  have  not  supposed  that  we  knew  anything  about  the 

formula  in  working  the  example. 

6.  Similar  figures1  vary  in  areas  as  the  squares  of  their  like  dimensions. 

The  diameter  of  one  circle  is  four  times  that  of  another.     Using  this 

principle,  find  the  relation  of  their  areas.     (Similar  figures  are  those  that 

are  alike  in  shape.) 

1  The  ideas  used  here  are  not  new,  but  will  help  to  show  the  intimate 
relation  between  variation  and  ratio  and  proportion.  (See  Art.  87.) 


344  PRACTICAL  MATHEMATICS 

Solution.     Here  the  relation  is  best  expressed  as  a  proportion.     Using 
A  and  a  for  the  areas,  and  4D  and  D  for  the  diameters,  we  have 


Dividing  both  terms  of  the  second  ratio  by  D1,  we  have 

A:a  =  16:l. 
/.  A  =  16a. 

6.  A  grindstone  when  new  is  48  in.  in  diameter.     How  large  is  it  in 
diameter  when    }   ground   away?     When   i   ground   away?     When    J 
ground  away?  Arts.  41.57—  in.;  33.94+  in.;  24  in. 

Suggestion.     Let  d\  in.  =  diameter  when  i  ground  away.     Then  since 
}  remains,  l:J=48J:di2.     Solving  for  di,  we  have  di=41.57  —  . 

7.  To  double  the  diameter  of  a  circle  has  what  effect  on  its  area?     To 
double  the  side  of  a  regular  hexagon  has  what  effect  on  its  area? 

Ans.  Multiplies  area  by  4. 

8.  Similar  solids  vary  in  volumes  as  the  cubes  of  their  like  dimensions. 
A  water  pail  that  is  10  in.  across  the  top  holds  12  quarts.     Find  the  vol- 
ume of  a  similar  pail  that  is  12  in.  across  the  top.     Ans.  20.736  quarts. 

9.  The  number  of  vibrations  made  by  a  pendulum  varies  inversely 
as  the  square  root  of  its  length.     A  pendulum  39.1  in.  long  makes  one 
vibration  per  second.     How  long  must  a  pendulum  be  to  make  four 
vibrations  per  second?     To  make  one  vibration  in  10  seconds? 

Ans.  2.44  in.  +  ;  325  ft.  10  in. 
k 
Solution.     The  law  is  n  =  ~7?>  where  n  =  number  of  vibrations  per 

Vl 

second,  and  1=  length  of  pendulum  in  inches. 
To  find  k,  put  n  =  l  and  J  =  39.1. 


To  find  the  length  of  a  pendulum  to  vibrate  four  times  per  second, 
put  n=4. 

\/3jU          ,   Z_39J_244  , 

To  find  the  length  of  a  pendulum  to  vibrate  once  in  10  seconds, 
put  n  =  iV 

and  /  =  100X39. 1=3910  in.  =325  ft.  10  in.  Ans. 


10       Vl 

10.  If  a  lever  with  a  weight  at  each  end  is  balanced  on  a  fulcrum,  the 
distances  of  the  two  weights  from  the  fulcrum  vary  inversely  as  the 
weights.     If  two  men  of  weights  160  Ib.  and  190  Ib.  respectively  are  bal- 
anced on  the  ends  of  a  10-ft.  stick,  what  is  the  length  from  the  fulcrum 
to  each  end?  Ans.  5*  and  4$  ft. 

11.  Could  a  1-lb.  weight  balance  a  100-lh.  weight?     How  could  they 
be  placed  on  a  3-ft.  bar? 

Ana.   Yes.     Fulcrum  2iVi  ft.  from  1-lb.  weight. 


VARIATION  345 


12.  If  /ocgg'  and/  oc--,  show  that/  <x  —  - 

13.  If  R  *  E,  R  oc  I,  and  R  <x  L  find  R  in  terms  of  E,  I,  and  d 


14.  If  the  illumination  of  an  object  varies  inversely  as  the  square  of 
the  distance  d  from  the  source  of  light,  and  the  illumination  of  an  object 
at  the  distance  8  ft.  from  the  source  of  light  is  3,  find  the  illumination  of 
an  object  at  32  ft.  from  the  source  of  light.     At  the  distance  of  20  ft.  from 
the  source  of  light.  Ans.   ^  ;  0.48. 

15.  The  number  of  units  of  heat  H,  generated  by  an  electric  current 
of  /  amperes  in  a  circuit,  varies  as  the  square  of  the  current  /,  as  the 
resistance  R,  and  as  the  time  t  in  seconds  during  which  the  current  passes. 
Write  the  law  in  the  two  forms.  Ans.  H  oc  727tt  •  H  =  kI2Rt. 

16.  By  trial  in  the  above  formula  it  is  found  that  H  =388,800  if  7  =  10, 
ft  =  9,    and   t  =  3Q  minutes.     Find   H  when  7  =  40,    72  =  50,    and  t  =  45 
minutes.  Ans.  77  =  51,840,000;  A:  =0.24. 

IT.  Two  parallelepipeds  ("rectangular  solids)  of  the  same  shape  have 
corresponding  dimensions  of  3^  ft.  and  1\  ft.  Find  the  relation  of  their 
volumes.  Ans.  343:3375. 

18.  A  circular  sheet  of  steel  2  ft.  in  diameter  increases  in  diameter  by 
755,  when  the  temperature  is  increased  by  a  certain  amount.     Find  the 
increase  in  area  of  the  sheet.  Ans.  TJySthr  or  nearly  T^. 

19.  A  wire  rope  1  in.  in  diameter  will  lift  10,000  Ib.     What  will  one 
f  in.  in  diameter  lift?  Ans.   1406  j  Ib. 

20.  A  cone  of  cast  iron  8  in.  high  weighs  50  Ib.     What  will  be  the 
weight  of  a  cone  of  the  same  shape  and  material  and  5  in.  high? 

Ans.  12.207+  Ib. 

21.  Two  persons  of  the  same  "build"  are  similar  in  shape.     A  man 
5^  ft.  tall  weighs  150  Ib.     Find  weight  of  a  man  of  same  "build"  and 
6ft.  tall.  Ans.   194.74+  Ib. 

22.  A  man  5  ft.  5  in.  tall  weighs  140  Ib.,  and  one  6  ft.  2  in.  tall  weighs 
216  Ib.     Which  is  of  the  stouter  "build"?          Ans.  The  216-lb.  man. 

23.  The  electrical  resistance  of  a  substance  varies  directly  as  the  length 
L,  and  inversely  as  the  area  A  of  the  cross  section.     If  the  resistance  of 
a  bar  of  annealed  aluminum  1  in.  long  and  1  sq.  in.  in  cross  section  is 
0.000001144  ohm  at  32°  Fahrenheit,  find  the  resistance  of  a  wire  of  .the 
same  material  1  ft.  long  and  0.001  in.  in  diameter. 

Ans.  17.48—  ohms. 
Note  that  you  have  in  this  exercise  the  law  of  Art.  253. 

24.  Find  the  resistance  of  1  mile  of  wire  ^  in.  in  diameter  of  above 
material  and  at  32°  F.  Ans.  94.5+  ohms. 

25.  If  the  resistance  of  a  coil  of  wire  of  the  above  material  at  32°  F. 
is  27.3  ohms,  and  the  wire  is  0.05  in.  in  diameter,  find  the  length. 

Ans.  3904.7ft, 

26.  If  the  resistance  of  a  wire  9363  ft.  long  is  21.6  ohms,  what  would  be 
the  resistance  if  its  length  were  reduced  to  5732  ft.  and  its  cross  section 
made  half  again  as  large?  Ans.  8.816—  ohms. 


340  PRACTICAL  MATHEMATICS 

27.  The  resistance  of  1  ft.  of  silver  wire  0.001  in.  in  diameter  at  32°  F. 
is  9.023  ohms.     Find  the  resistance  of  a  silver  wire  1  meter  long  and  1 
mm.  in  diameter.  Ana.  0.0191  ohm. 

Suggestion.     Use  0.001  in.  =0.0254  mm.  and  1  ft.  =0.3048  m. 

28.  The  resistance  of  1  meter  of  German  silver  wire  1  mm.  in  diameter 
is  0.2659  ohm.     Find  the  resistance  of  1  ft.  of  wire  of  the  same  material 
and  0.001  in.  in  diameter.  Ans.   125.6+  ohms. 

29.  Find  the  resistance  of  a  coil  of  copper  wire  0.03  in.  in  diameter, 
the  coil  being  18  in.  in  diameter  and  having  300  turns,  if  1  ft.  of  copper 
wire  0.001  in.  in  diameter  has  a  resistance  of  9.803  ohms. 

Ans.  15.4—  ohms. 

30.  There  are  three  wires  of  diameters  2  mils,  3  mils,  and  4  mils  respec- 
tively.    What  length  must  the  second  and  third  have  to  have  the  same 
resistance  as  20  ft.  of  the  first?  Am.  45  ft.  and  80  ft. 

31.  The  size  of  a  stone  carried  by  a  swiftly  flowing  stream  varies  as 
the  sixth  power  of  the  velocity  of  the  water.     If  the  velocity  of  a  stream 
is  doubled,  what  effect  does  it  have  on  its  carrying  power?     What  effect 
if  changed  from  5  miles  per  hour  to  15  miles  per  hour? 

Ans.  Multiplies  by  64.     Multiplies  by  729. 

282.  Transverse  strength  of  wooden  beams. — Other  things 
being  equal,  the  strength  of  a  beam,  rectangular  in  cross  section, 
and  supported  at  each  end,  varies  (1)  inversely  as  the  length 
in  feet,  (2)  directly  as  the  breadth  in  inches,  and  (3)  directly 
as  the  square  of  the  depth  in  inches. 

6 


Fiu.  200. 

That  the  strength  varies  inversely  as  the  length  in  feet 
means  that  if  a  beam,  supported  horizontally  as  in  Fig.  206, 
has  its  length  increased  and  everything  else  unchanged,  the 
weight  that  it  will  support  is  decreased  in  the  same  ratio 
as  the  length  was  increased.  That  is,  if  the  length  is  doubled 
it  will  support  one-half  as  great  a  weight.  If  W  is  the  weight 
and  L  the  length,  in  the  language  of  variation  this  fact  is 
stated  thus, 


VARIATION  347 

The  length  L  is  the  distance  between  the  points  of  support. 

That  the  strength  varies  directly  as  the  breadth  in  inches 
means  that  if  the  breadth  b  is  increased  in  a  certain  ratio 
and  everything  else  is  unchanged,  the  weight  that  it  will 
support  is  increased  in  the  same  ratio.  That  is,  if  the  breadth 
is  doubled  the  weight  that  it  will  support  will  be  doubled. 
This  law  is  expressed  in  symbols  thus, 

W  a  6,  or  W  =  kj). 

That  the  strength  varies  directly  as  the  square  of  the  depth 
in  inches  means  that  if  the  depth  d  is  increased  in  a  certain 
ratio  while  other  things  remain  the  same,  the  weight  that  it 
will  support  is  increased  by  the  square  of  the  ratio  of  that 
increase.  That  is,  if  the  depth  is  doubled  the  weight  that  it 
will  support  is  four  times  as  great.  In  symbols  this  is  ex- 
pressed thus, 

W  a  d2,  or  W  =  ksd2. 

Finally,  if  the  length,  breadth,  and  depth  are  all  changed  we 
have  a  combination  of  all  these  laws,  and  it  may  be  expressed 
thus  (see  Art.  281), 

bd*  Kbd2 

TT  oc-y-,  or  W  =  — j — 
Li  LJ 

By  the  expression  "other  things  being  equal"  is  meant 
that  the  material  must  be  the  same,  and  the  beams  must  be 
similarly  supported  and  similarly  loaded.  The  nature  of  the 
timber  is  an  important  factor,  since  timber,  even  of  the  same 
kind,  varies  in  strength  to  a  considerable  extent.  Each 
beam  therefore  has  what  is  called  a  natural  constant,  the  K 
of  the  formula,  which  must  be  considered  in  the  calculation 
of  its  carrying  capacity.  This  constant  is  the  same  for  beams 
when  these  "other  things"  are  equal.  That  is,  for  beams 
of  exactly  the  same  material,  supported  the  same  and  weighted 
the  same. 

283.  The  constant. — To  find  this  constant,  it  is  usual  to 
take  a  bar  of  similar  material,  1  in.  square  in  cross  section,  and 
long  enough  to  allow  of  its  being  placed  on  supports  1  ft.  apart. 
The  constant  to  be  used  with  beams  weighted  in  the  middle 


348  PRACTICAL  MATHEMATICS 

is  the  weight  of  the  central  load  which  is  just  sufficient  to 
break  the  bar.  The  constant  may  be  expressed  in  pounds, 
hundred  weight,  tons,  etc.,  and  the  carrying  capacity  of  the 
beam  it  is  applied  to  will  always  be  in  the  same  units. 

284.  Factor  of  safety. — For  the  man  who  wishes  to  apply 
these  facts,  another  important  consideration  is  the  ratio  which 
the  breaking  load  bears  to  the  safe  load.  This  ratio  is  called  the 
factor  of  safety.  Its  value  depends  upon  whether  the  load  is 
a  live  load,  that  is,  a  moving  load;  or  a  dead  load,  that  is,  a 
stationary  load. 

The  factor  of  safety  for  a  dead  load  is  usually  taken  as  5, 
which  means  that  the  safe  load  upon  a  beam  must  not  be  more 
than  one-fifth  of  the  breaking  load.  The  factor  of  safety  for 
a  live  load  is  often  taken  as  10. 


EXERCISES  100 

Kbd* 

1.  Solve  the  formula  W  =  — = —  for  each  letter  used. 
Li 

_Kbd*          WL  IWL          TTL 

W  '  b~Kd*'  l       \AV          6d* 

2.  The  constant  for  white  pine  is  300  pounds.     Find  the  weight  a 
beam  of  this  material,  centrally  loaded,  and  10  ft.  long,  3  in.  broad,  and 
7  in.  deep,  will  support.     How  much  is  the  safe  load  if  the  factor  of  safety 
is  5?  Ans.  44101b.;  882  Ib. 

3.  How  long  may  a  beam  of  white  pine,  centrally  loaded,  be  between 
supports,  if  it  is  to  have  a  safe  load  of  750  Ib.,  and  is  3  in.  by  8  in.  set  on 
edge?  Ans.   15.36ft. 

4.  By  experiment  we  find  that  a  beam  of  pine  40  ft.  long,  1  ft.  broad, 
and  1  ft.  deep,  will  carry  a  load  of  4500  Ib.     Find  the  depth  of  a  beam  of 
the  same  wood  similarly  loaded  to  carry  a  load  of  1200  Ib.,  when  the 
length  is  6  ft.  and  the  breadth  is  2  in.  Ans.  5.88—  in. 

T'l     Jj 

Solution.     Use  the  formula  W  =  — j — ,  and  substitute  W  =4500,  L  =40, 

ftT  1  9  1  *^2  A*?^ 

6  =  12,  and  d  =  l2.     This  gives  4500  =  — ^ .'.  A'  =  -g- 

625 
Substituting   JK  =  1200,  L=6,  6=2,  and  A'  =  -g    in  the  same  formula, 

1200  =  — Q~ 

'aolving  for  <l,  r/  =  5.88-. 

6.  In  the  above  exercise  find  the  depth  if  we  allow  a  factor  of  safety 
of  10.  Ans.  18.6-  in. 


VARIATION  349 

6.  In  beams  having  the  load  uniformly  distributed,  the  constant  is 
twice  as  large  as  when  the  load  is  centrally  located.     Solve  exercise  2  if 
the  beam  is  uniformly  loaded.  Ans.  8820  lb.;  1764  Ib. 

7.  Find  the  breadth  of  a  beam  of  oak,  resting  upon  supports  18  ft. 
apart,  the  beam  being  12  in.  deep,  to  carry  safely  a  uniformly  distributed 
load  of  5  tons.     The  constant  is  500  lb.  if  loaded  in  center,  and  the  factor 
of  safety  is  5.  Ans.  6j  in. 

8.  A  hall  16  ft.  wide  has  the  floor  supported  by  joists  of  pine  3  in.  by 
12  in.  set  on  edge;  using  the  constant  300  lb.  and  a  factor  of  safety  of 
5,  find  how  far  the  joists  must  be  placed  from  center  to  center  to  support 
a  load  of  140  lb.  per  square  foot  of  floor  surface.    Ans.   17f  in.  nearly. 

9.  O'Connor  gives  the  following  formula  to  calculate  the  dead  dis- 
tributed safe  load  on  timber,  supported  at  both  ends,  and  of  rectangular 
cross  section  (this  includes  floor  joists): 

4bd*K' 
•-2L-' 
where   W=  load  in  pounds, 

b  =  breadth  in  inches, 

d  =  depth  in  inches, 

L=  length  of  span  in  inches, 

K  =  1900  for  oak  and  1100  for  fir. 

(Notice  how  this  formula  differs  from  the  one  previously  given.)  What 
safe  weight  distributed  will  an  oak  beam  6  in.  by  10  in.,  set  on  edge,  sup- 
port if  the  span  is  16  ft.?  Ans.  11,875  lb. 

10.  The  joists  in  a  room  14  ft.  wide  and  26  ft.  long  are  fir  3  in.  by  10  in. 
How  far  should  their  centers  be  placed  apart  if  the  floor  is  to  support  a 
crowd  of  men?     (A  crowd  of  men  closely  packed  average  140  lb.  per 
square  foot.)  Ans.  2  ft.  nearly. 

11.  From  another  source  the  following  is  taken:  For  white  pine  beams 

.     2000K  bd2     .  ,     , 

the  formula  W  =    0    X  -j-  gives  the  safe  load  when  the  beam  is  sup- 

O  Ll 

ported  at  both  ends  and  loaded  in  the  middle. 

TF  =  safe  load  in  pounds,  less  weight  of  beam, 
L=  length  of  beam  in  inches, 
d  =  depth  of  beam  in  inches, 
b  =  breadth  of  beam  in  inches. 

(1)  Given  L  =  12  ft.,  6=3  in.,  and    rf  =  8  in.,  find  W. 

(2)  Given  L  =  12  ft.,  b=  8  in.,  and    d  =  3  in.,  find  W. 

(3)  Given  L  =  16  ft.,  b  =   4  in.,  and  W  =  1900,  find  d. 

(4)  Given  6=6  in.,  d  =  W  in.,  and  TF  =  4100,  find  L. 

Ans.   (1)  889  lb.;  (2)  333|  lb.;  (3)  11.7-  in.;  (4)  8  ft.  nearly. 

12.  The  following  are  the  results  of  some  experiments  on  the  strength 
of  timbers  when  loaded  in  the  middle.      In  each  case  find  the  strength  of  a 
stick  of  the  same  material  1  ft.  long,  1  in.  in  breadth,  and  1  \n.  in  depth. 
This  value  is  called  K  in  the  list. 


350 


1'RACTICAL  MA  THEM  A  TICS 


Name  of  wood 

Length 
in 
feet 

Breadth 
in 
inchi-a 

Depth 
in 
inches 

Breaking 
weight  in 
pounds 

A" 

White  pine  

2 

2 

2 

1,430 

3574 

Yellow  pine  

10$ 

14 

15 

68,000 

232  + 

Pitch  pine  

10} 

14 

15 

118,500 

404.4 

Ash  

2 

2 

2 

2,052 

513 

Pitch  pine  

7 

2 

2 

622 

544J 

Ash  

7 

2 

2 

772 

675  J 

Fir                      

7 

2 

2 

420 

367  i 

13.  Suppose  there  are  three  pieces  of  timber  of  the  following  dimen- 
sions: 

(1)  12  ft.  long,  6  in.  deep,  and  3  in.  thick; 

(2)  8  ft.  long,  5  in.  deep,  and  4  in.  thick; 

(3)  15  ft.  long,  9  in.  deep,  and  8  in.  thick. 

Compare  their  strengths.  Ans.  In  the  ratio  of  9: 12.5:43.2. 

14.  Given  a  stick  of  timber  14  ft.  long,  8  in.  deep,  and  3  in.  thick;  find 
the  depth  of  another  piece  of  the  same  material  18  ft.  long  and  4  in. 
thick  that  will  support  five  times  as  much  as  the  first.   Ans.  17.6—  in. 

Solution.     Let  W  =  weight  first  stick  will  support. 

Kbd*    K  3-82    96 A' 


Then      W 


14 


In  the  case  of  the  second  stick  d  is  to  be  found  when  5W  =  --  _—  is  the 


weight. 


480A' 
7 


K  4-d* 
18 


Solving,  d  =  17.6-. 

15.  Given  a  piece  of  timber  12  ft.  long,  6  in.  deep,  and  4  in.  thick; 
find  the  thickness  of  another  stick  of  the  same  material  16  ft.  long  and 
8  in.  deep  that  will  support  twice  as  much  as  the  first.  Ans.  6  in. 

16.  Given  a  stick  of  timber  12  ft.  long,  5  in.  deep,  and  3  in.  thick; 
find  the  thickness  of  another  stick  of  the  same  material  14  ft.  long  and 
0  in.  deep  that  will  support  four  times  as  much  as  the  first. 

Ans.  9.72+  in. 

17.  Given  a  stick  of  timber  12  ft.  long,  6  in.  deep,  and  4  in.  thick; 
find  the  depth  of  a  stick  of  the  same  material  20  ft.  long  and  5  in.  thick 
that  will  support  twice  the  weight  of  the  first.  Ans.  9.8—  in. 


CHAPTER  XXXIII 
GRAPHICS 

285.  The  graph. — The  temperatures  read  each  hour  during 
March  21,  1917,  were  as  given  in  the  following  table : 


Hour,  A.  M  

1? 

1 

2 

8 

4 

5 

6 

7 

8 

q 

10 

11 

^^> 

Temperature 

45 

45 

45 

45 

43 

4? 

41 

40 

4? 

51 

57 

5q 

6? 

Hour   P.  M 

1 

2 

3 

4 

5 

6 

7 

8 

q 

10 

11 

1? 

Temperature  

66 

70 

74 

76 

76 

75 

74 

73 

72 

70 

69 

68 

The  change  in  temperature  is  quite  easily  seen  from  a  study 
of  this  table,  but  it  may  be  seen  at  a  glance  if  we  put  the  facts 
into  a  diagram  as  shown  in  Fig.  207.  Here  the  hours  are 
located  on  a  horizontal  line  and  the  degrees  on  a  line  perpen- 
dicular to  it.  The  reading  for  any  hour  is  located  so  as  to  be 
above  the  hour  and  to  the  right  of  the  degree  of  temperature. 
In  this  manner  we  can  locate  24  points. 

Evidently  if  the  temperature  reading  had  been  taken  each 
minute  instead  of  each  hour,  there  would  be  determined  60 
points  in  the  same  space  in  which  we  now  have  one.  If  we 
suppose  no  sudden  change  in  temperature  between  the  hourly 
readings,  we  may  connect  the  points  representing  these  by  a 
line,  and  any  point  on  this  line  will  indicate  the  temperature 
for  the  corresponding  time.  Of  course,  if  there  had  been  a 
sudden  rise  and  fall  in  temperature  between  the  hours,  it  is  not 
shown  by  this  line.  The  oftener  then  we  take  the  readings, 
the  truer  will  the  line  indicate  the  changes.1 

The  representation,  made  as  in  Fig.  207,  is  called  a  graph; 

1  Cross-ruled  paper,  or  paper  ruled  into  squares  of  various  sizes,  can 
be  obtained  cheaply ;  and,  by  using  this  paper,  the  points  can  be  located 
with  greater  accuracy  than  on  plain  paper. 

351 


352 


PRACTICAL  MATHEMATICS 


and  such  a  method  of  representing  relations  between  numbers 
is  called  a  graphical  method. 

The  use  of  the  graph  is  of  very  wide  application.  At 
the  weather  bureaus  there  are  thermometers  with  an  attach- 
ment which  automatically  traces  the  graph  of  the  temperature 
and  time.  Engineers  make  constant  use  of  the  graph  in  their 


40    - 


35 

12128456789  10  11121    88456    789  10  11  12 
A.M.  Hours  P.M. 

March.  21,  1917 
FlO.  207. 

work.  Laboratory  data  are  put  in  the  form  of  a  graph. 
Graphs  can  be  made  for  algebraic  equations,  and  relations  are 
thus  clearly  shown  that  otherwise  would  be  difficult  to  see. 

Making  the  graph  is  often  spoken  of  as  plotting. 

286.  Definitions  and  terms  used. — Since  we  often  wish 
to  plot  negative  as  well  as  positive  numbers,  it  is  necessary 
to  give  certain  definitions  and  make  certain  assumptions  which 
will  now  be  explained. 


GRAPHICS 


353 


If  in  Fig.  208  OX  and  OY  are  drawn  at  right  angles  to 
each  other,  the  position  of  any  point,  as  P,  may  be  located  by 
measuring  its  distance  from  OY  and  from  OX.  These  lengths, 
which  in  the  figure  are  OA  and  OB,  are  called  the  coordinates 
of  the  point  P.  The  length  OA  is  called  the  x-cob'rdinate ; 
and  OB,  the  y-coordinate. 

The  two  lines  OX  and  OY  are  called  the  x-axis  and  the 
y-axis  respectively.  Together  they  are  spoken  of  as  the 
coordinate  axes.  The  point  0  where  the  two  axes  cross  is 
called  the  origin. 


II 


III 


o 


-X 


IV 


FIG.  208. 

The  rr-axis  is  called  the  axis  of  abscissas,  and  the  y-axis, 
the  axis  of  ordinates.  The  ^-coordinate  and  the  ^-coordinate 
are  also  called  the  abscissa  and  ordinate  respectively  of  the 
point. 

The  coordinates  are  always  measured  from  the  origin.  Any 
abscissa  measured  toward  the  right  is  positive,  and  measured 
toward  the  left  is  negative.  Any  ordinate  measured  upward 
is  positive  and  downward  is  negative. 

The  four  parts  into  which  the  axes  divide  the  plane  are 
called  quadrants.  These  are  called  the  first,  second,  third, 
and  fourth  quadrants,  and  are  numbered  in  Fig.  208  by  the 
numerals  I,  II,  III,  IV. 

It  is  evident  that,  in  the  first  quadrant,  both  coordinates 
are  positive;  in  the  second  quadrant,  the  abscissa  is  negative  and 


354 


PRACTICAL  MA  THEM  A  TICS 


the  ordinate  positive;  in  the  third  quadrant,  both  coordinates 
are  negative;  in  the  fourth  quadrant,  the  abscissa  is  positive 
and  the  ordinate  is  negative.  This  is  shown  in  the  following 
table: 


Quadrant  

I 

II 

III             IV 

Abscissa  ... 

+ 

•f 

Ordinate. 

, 

+ 

-       1       - 

287.  Plotting  points. — To  plot  a  point  is  to  locate  it  with 
reference  to  a  set  of  coordinate  axes.     To  plot  a  point  whose 


y 

P 

(-2 

,5) 

r> 

P 

(  J 

,4} 

tt 

0 

_y 

/ 

p 

(-4 

_  «• 
>     * 

I) 

P 

f4' 

-a 

) 

FIG.  209. 

x-coordinate  is  3  and  y-coordinate  4,  first  draw  the  axes  (see 
Fig.  209),  then  choose  a  unit  of  measure  and  lay  off  OA  3 
units  to  the  right  of  0.  Through  A  draw  a  line  parallel  to  the 
t/-axis.  Now  lay  off  OB,  4  units  above  0,  and  through  B 
draw  a  line  parallel  to  the  z-axis.  The  required  point  is 
located  where  these  two  lines  meet.  In  the  figure  it  is  located 
as  P(3,  4),  which  is  the  usual  manner  of  writing  the  coordinates 


GRAPHICS 


355 


of  a  point.  The  abscissa  is  placed  first;  they  are  separated  by 
a  comma  and  inclosed  in  parentheses.  It  is  read:  "The 
point  P  whose  coordinates  are  3  and  4." 

In  a  similar  manner  the  following  points  are  located  as 
given  in  the  figure:  P(-2,  5),  P(-4,  -3),  P(4,-5). 


3 

§• 

PH 


/ 

/ 

/ 

/ 

j 

/ 

20 

/ 

/ 

z 

/ 

/ 

/ 

/ 

j* 

X 

1C. 

^ 

LC 

0" 

y 

Ib 

50 

G 

0 

7 

0    7 

3    b 

0 

j 

0 

Ye 

0 

irs 

) 

1 

0 

2 

0 

3 

0 

FIG.  210. 


EXERCISES  101 

1.  Plot  the  following  points:  (-1,  6),  (7-2),  (0,  4),  (4,  0),  (-6,  0), 
(-3,  -8),  (-6,2),  (0,0). 

2.  The  temperatures  read  each  hour  for  the  24  hours  ending  Jan. 
26,  1918,  at  2  p.  M.  were  as  follows:  Above  zero  11,  11,  10,  6,  8,  5,  4,  3,  2, 
2,  1,  0,  0;  below  zero  1,  1,  2,  3,  2;  above  zero  1,  3,  6,  8,  10,  12.     Plot  these 
using  the  hours   as  abscissas  and  the  temperatures  as  ordinates,  and 
connect  by  a  smooth  curve. 

3.  The  population  of  a  town  in  1850  was  16,000;  in  1860,  16,300;  in 
1870,  16,850;  in  1880,  17,800;  in  1890,  19,100;  and  in  1900,  20,700.     Plot 
a  curve  showing  the  variation  in  population,  and  estimate  the  probable 
population  in  1875  and  in  1907. 

Discussion.  Draw  the  axes;  lay  off  the  years,  beginning  with  1850  at 
the  origin,  along  the  x-axis ;  and  lay  off  the  population  in  thousands  along 
the  y-axis,  beginning  with  16,000  at  the  origin.  The  curve  will  then  be 
as  in  Fig.  210. 

The  population  for  any  year  is  estimated  by  locating  the  year  on  the 
x-axis;  drawing  a  perpendicular  to  the  x-axis;  and,  from  the  intersection 
of  this  with  the  curve,  drawing  a  parallel  to  the  x-axis.  The  point  where 
this  parallel  intersects  the  y-axis  determines  the  population. 


356 


PRACTICAL  MATHEMATICS 


The  process  by  which  we  determine  the  coordinates  of  any  point  on 
the  curve  is  called  interpolating. 

4.  A  company  began  to  sink  a  mining  shaft  on  June  4;  on  June  11  it 
had  reached  a  depth  of  30  ft. ;  on  June  18,  54  ft. ;  on  June  25,  70  ft. ;  on 
July  2,  82  ft. ;  on  July  9,  100  ft. ;  on  July  16,  135  ft.,  and  on  July  23,  150ft. 
Plot  a  curve  showing  the  progress  of  the  work. 

5.  The  population  of  the  United  States  by  decades  was  as  follows. 
Plot  and  estimate  the  population  for  1913  and  1919. 


Yrar 

Population 

Year 

Population 

1790 

3,929,214 

1860 

31,443,321 

1800 

5,308,433 

1870 

38,558,371 

1810 

7,229,881 

1880 

50,155,783 

1820 

9,663,822 

1890 

62,669,756 

1830 

12,806,020 

1900        76,295,200 

1840 

17,069,453 

1910 

91,972,266 

1850 

23,191,876 

6.  Make  a  graph  from  which  can  be  read  the  product  of  any  number 
from  -225  to  225  multiplied  by  0.367. 

Suggestion.  Plot  the  numbers  from  —225  to  225  as  abscissas  and  the 
products  as  ordinates. 


•3 

o  4 
£3 

0 

S  2 
H 

1 

0 

-J— 

A' 

_—  •  -• 

—  • 

_—  - 

g- 

-rr 

—  —  ' 

_^—  - 

7~ 

•-- 

-- 

,_, 

— 

— 

^-, 

-jT 

^  —  ^ 

^ 

^-^ 

^ 

^° 

, 

/° 

f 

A 

p 

10  20  30  40  60              80        100                        150 

Distance  in  Feet 
Fio.  211. 

7.  Make  a  graph  from  which  can  be  read  the  quotient  of  any  number 
less  in  absolute  value  than  500  divided  by  12.7. 

8.  A  stone,  falling  from  rest,  falls  through  4  ft.  in  J  second,  16  ft.  in  1 
second,  36  ft.  in  1 J  seconds,  64  ft.  in  2  seconds,  100  ft.  in  2§  seconds,  and 
144  ft.  in  3  seconds.     Find,  by  plotting  a  curve,  how  long  a  stone  would 
be  in  falling  80  ft. ;  and  also  how  far  it  falls  in  2}  seconds. 


GRAPHICS 


357 


The  graph  for  exercise  8  is  as  shown  in  Fig.  211.  To  find  how  long  the 
stone  would  be  in  falling  80  ft.,  locate  80  ft.  on  the  x-axis,  draw  a  per- 
pendicular line  AB,  and  through  the  point  B  on  the  curve  draw  a  line 
parallel  to  the  x-axis  to  intersect  the  y-axis.  This  point  of  intersection 
C  determines  the  number  of  seconds  it  will  take  the  stone  to  fall  80  ft. 
To  find  how  far  the  stone  will  fall  in  2f  seconds,  proceed  in  a  similar  man- 
ner, starting  on  the  y-axis. 

9.  The  football  accidents  for  the  years  given  are  as  follows: 


Year 

Deaths 

Injuries 

Year 

Deaths 

Injuries 

1901 

7 

74 

1910 

22 

499 

1902 

15 

106 

1911 

11 

178 

1903 

14 

63 

1912 

13 

1904 

14 

276 

1913 

14 

1905 

24 

200 

1914 

12 

1906 

14 

160 

1915 

16 

1907 

15 

166 

1916 

19 

1908 

11 

304 

1917 

12 

1909 

30 

216 

Plot  two  curves,  using  the  years  as  abscissas  and  the  deaths  and  injuries 
respectively  as  ordinates.  State  the  conclusions  that  you  can  draw  from 
the  curves. 

10.  The  monthly  wages  of  a  man  for  each  of  his  first  13  years  of  work 
was  as  follows:  $28,  $30,  $37.50,  $45,  $60,  $65,  $90,  $95,  $95,  $137, 
$162,  $190,  and  $210. 

Plot  the  curve  showing  the  change.  Estimate  his  salary  for  the  four- 
teenth and  fifteenth  years.  Can  you  be  certain  of  his  salaries  for  these 
years?  Why? 

11.  A  cyclist  starts  from  a  town  at  8  A.  M.  and  rides  4  hours  at  the 
uniform  rate  of  9  miles  per  hour.     He  then  rests  1  hour  and  returns  at 
the  rate  of  8  miles  per  hour.     At  11  A.  M.  a  second  cyclist  starts  from  the 
same  town  and  rides  over  the  same  route  at  the  rate  of  6  miles  per  hour. 
Plot  curves  showing  where  they  will  meet. 

12.  (a)   Reckon  the  simple  interest  at  6%  on  $100  for  the  several 
years  1,  2,  3,   •  •  •  10.      Plot  the  years  as  abscissas  and  the  interest  as 
ordinates.     Connect  with  a  smooth  curve  from  which  may  be  read  the 
interest  on  $100  at  6%  for  any  time. 

(b)  Reckon  the  compound  interest  at  6%  on  $100  for  the  several 
years  1,  2,  3,  •  •  •  10.  Plot  as  in  (a),  using  the  same  axes. 

13.  Plot  the  numbers  1,  4,  9,  16,  25,  36,  etc.,  as  abscissas  and  their 
square  roots  as  corresponding  ordinates.     Make  scale  for  ordinates  5 
or  10  times  that  of  abscissas.     How  can  you  determine  roots  of  interven- 
ing numbers?     (See  discussion  of  exercise  3.)     Find  roots  of  11,  14,  21, 


PRACTICAL  MATHEMATICS 


29,  42  from  curve.     How  do  these  agree  with  computed  values  of  roots 
of  these  numbers? 

14.  If  T  is  the  tensile  strength,  in  tons  per  square  inch  of  cross  section, 
of  steel  containing  X  per  cent  of  carbon,  and  we  are  given  the  following 
values,  plot  a  curve  to  show,  as  accurately  as  the  data  will  allow,  the 
tensile  strength  of  steel  containing  any  percentage  of  carbon  from  0. 1  to 
1  per  cent.  What  strength  would  you  expect  for  0.4  per  cent  of  carbon? 


X 

0.14 

0.46 

0.57 

0.66 

0.78 

0.80 

0.87 

0  96 

T 

28.1 

33.8 

35.6 

40 

41.1 

45.9 

46.7 

52.7 

16.  Plot  the  Fahrenheit  thermometer  scale  on  the  z-axis  and  the  centi- 
grade scale  on  the  y-axis.  Draw  the  line  any  point  of  which  has  as  co- 
ordinates equivalent  temperatures  on  the  two  scales.  How  can  this 
graph  be  used  to  reduce  from  one  scale  to  the  other?  Read  from  the 
graph  the  centigrade  temperatures  for  the  following  Fahrenheit  readings: 
25°,  90°,  367°,  -40°,  15°.  Give  in  Fahrenheit  the  following  centigrade 
readings:  33°,  76°,  15°,  -46°,  -9°. 

16.  In  an  experiment  on  the  stretching  of  an  iron  rod  the  linear  exten- 
sion, L  in  inches,  for  a  load,  W  in  pounds,  was  found  to  be  as  follows: 


W.         .600 

1100 

1600 

2100    2600 

3100 

3600    4100 

4600   5100 

) 

i 

I 

n    nt\r\ 

n    rn  o 

n    r\OT 

Plot,  choosing  suitable  distances  on  the  z-axis  for  W,  and  on  y-axis  for 
L.  Up  to  how  great  a  load  is  the  extension  proportional  to  the  load? 
That  is,  where  does  the  curve  change  direction  rapidly? 

Remark.  While  the  curve  is  a  straight  line,  the  relation  between  L 
and  W  can  be  expressed  in  the  language  of  variation  by  the  equation 
W  =  kL.  Or  we  may  say  that  the  straight  line  is  the  graph  of  a  direct 
variation  where  the  ratio  is  constant. 

17.  A  stick  of  white  pine  1  in.  broad  and  0.5  in.  thick  is  supported  at 
points  24  in.  apart  and  loaded  in  the  middle.  The  deflection,  d  in  inches, 
for  a  load,  W  in  pounds,  is  as  follows: 


W 

o 

5 

8 

18 

?8 

38 

48 

58 

63 

68  j  69 

70 

d 

o 

0088 

0  14 

035 

056 

077 

099 

1  ?? 

1  35 

1  685  1  765 

1  85 

1 

Plot  so  as  to  show  the  deflection  is  proportional  to  the  weight  up  to  a 
certain  point. 

18.  Plot  the  following  data  of  the  weight  of  railway  locomotives.  The 
weights  are  of  the  largest  locomotives  made  each  year.  Plot  time  on 
the  x-axis. 


GRAPHICS 


359 


Year 

1898 
1899 
1900 
1902 
1904 
1905 
1909 
1910 
1911 


Engine,  pounds 

230,000 
232,000 
250,300 
259,800 
287,240 
334,000 
425,900 
440,000 
616,000 


Engine  and  tender, 
pounds 
334,000 
364,000 
391,400 
383,800 
453,000 
477,000 
.  596,000 
611,800 
850,000 


19.  A  rifle  sighted  to  1000  yd.  rests  upon  a  support  5  ft.  from  the 
ground  and  is  fired.  The  height  of  the  bullet  above  the  support  at  the 
various  distances  is  given  in  the  following  table: 


Distance  in  yards   1 

100 

200 

300 

400 

500 

600 

700 

800 

900 

1000 

from  firing  point.    J 

Vertical    height        1 

above  support        \ 

7.3 

11.2 

15.0 

18.5 

21.0 

23.3 

25.0 

22.5 

16.5 

0 

in  feet. 

Plot  a  representation  of  the  path  of  the  bullet.  Show  the  ground  level 
and  the  height  of  the  support.  Where  does  the  bullet  reach  a  height  of 
20  ft.  ? 

20.  Plot  the  relation  between  centimeters  and  inches. 

Suggestion.  Take  the  numbers  of  centimeters  on  the  z-axis  and  the 
inches  on  the  y-axis.  Any  point  on  the  curve  has  as  coordinates  a  num- 
ber of  centimeters  and  a  number  of  inches  that  are  equivalent  in  length. 

21.  Plot  a  curve  showing  the  relation  between  yards  and  meters. 

22.  Plot  a  curve  showing  the  relation  between  pounds  and  kilos. 

23.  Plot  a  curve  showing  the  relation  between  dollars  and  francs. 
Use  1  franc  =  19.3  cents. 

288.  Graph  of  an  equation. — If  we  have  an  equation  in  two 
unknowns,  as  3x+4y  =  l2,  we  can  determine  a  number  of 
pairs  of  values  for  x  and  y  that  will  satisfy  the  equation. 
(See  Art  255.)  If  we  consider  each  of  these  pairs  as  the  co- 
ordinates of  a  point,  the  value  of  x  for  the  abscissa  and  y  for 
the  ordinate,  then  the  graph  determined  by  these  points  is 
called  the  graph  of  the  equation  or  the  curve  of  the  equation. 

To  plot  the  curve  of  3z+4?/  =  12,  determine  the  pairs  of 
values:  (0,  3),  (4,  0),  (2,  1J),  (3,  f),(5,  -f),  (8,  -3),  (-8,  9), 


360 


I'KACTH  'AL  MA  Til  KM  A  TICS 


(—4,  6),  (12,  —6).    Plot  these  points  and  connect  with  a 
smooth  line.     The  curve  is  shown  in  Fig.  212. 

Whenever  the  equation  is  of  the  first  degree  the  graph  will 
be  a  straight  line.  Since  a  straight  line  is  determined  by 
knowing  two  points  on  it,  the  graph  can  be  drawn  when  two 
points  only  have  been  plotted.  Usually  the  most  convenient 
points  to  take  are  those  where  the  line  crosses  the  two  axes. 
These  two  points  are  found  by  putting  x  =  0  and  finding  y,  and 


r 

A' 

s 

^ 

s 

X, 

f- 

8. 

9 

N 

^ 

s 

X, 

t(. 

4 

g 

x 

S, 

s 

s, 

« 

), 

\) 

s 

(' 

(3 

e.> 

,(-. 

) 

s 

S 

,1 

. 

» 

II 

S 

»'. 

' 

\ 

s 

s, 

vf 

}.- 

i) 

% 

*s, 

s 

s 

1 

-f 

| 

s 

x 

s 

S 

N 

V, 

Fio.  212. 


then  putting  y  =  0  and  finding  x.  The  distances  from  the 
origin  to  these  points  are  called  the  intercepts  on  the  axes. 
If  the  line  goes  through  the  origin  another  point  will  be  nec- 
essary. If  the  two  points  are  near  together  the  line  will  not 
be  well  determined  unless  the  work  is  very  accurate. 

289.  Simultaneous  equations.  —  If  we  plot  the  graphs  of 
two  simultaneous  equations  they  will  intersect  at  some  point 
if  they  are  not  parallel.  The  coordinates  of  the  point  of 
intersection  evidently  satisfy  both  equations  and  are  the  values 
obtained  by  solving  the  equations  as  simultaneous. 

Example.  Plot  each  of  the  following  equations  and  solve 
them  as  simultaneous  equations. 


(1) 


(2)  3z+2y  = 


GRAPHICS 


301 


Pairs  of  values  for  first:  (0,  £),  (1,  |),  (3,  0),  (7,  -1). 
Pairs  of  values  for  second:  (0,  2),  (2,  -1),  (4,  -4),  (-2,  5). 
These  are  plotted  and  the  intersection  is  determined  to  be  the 
point  (1,  |).  Solving  as  simultaneous  equations,  we  find  re  =  1 
and  y  =  \  as  values  of  x  and  y.  The  plotting  is  shown  in 
Fig.  213. 


FIG.  213. 

290.  The  graph  of  an  equation  of  any  degree.  —  The  grapt 
of  an  equation  of  any  degree  in  two  variables  (unknowns' 
can  be  plotted  by  taking  values  for  one  variable  and  deter- 
mining corresponding  values  for  the  other  variable.  Each  pair 
of  values  determines  a  point,  and  a  sufficient  number  of  these 
points  will  determine  the  form  of  the  curve. 

A  graph  of  an  equation  of  higher  degree  than  the  first  is 
not  a  straight  line.  The  graph  of  an  equation  of  the  second 
degree  in  two  variables  is  a  conic  section,  that  is,  the  section 
of  a  cone. 

EXERCISES  102 

Plot  and  solve  the  following  systems  of  equations: 


=  ll  and 


and 


3(52 


ritACTH  'AL  MA  THEM  A  TICS 


6.  Plot  the  curve  of  the  equation  xy  —  1.     This  will 
give  a  curve  from  which  can  be  read  the  reciprocals  of 
any  number.     It  is  plotted  by  first  finding  a  number 
of  pairs  of  values  for  z  and  y  which  satisfy  the  equa- 
tion, and  then  plotting  the  points  that  have  these 
pairs  as  coordinates.     It  should  be  noticed  that  when 
x  is  negative,  y  is  negative,  and  when  x  is  positive,  y 
is  positive. 

The  pairs  of  values  given  m  the  table  are  found  and 
plotted  as  shown  in  Fig.  214. 

7.  Plot  the  curve  for  j/  =  x*,  and  thus  find  the  curve 
from  which  can  be  read  squares  and  square  roots  of 
numbers. 


Pairs  of  values 

x              y 

±A 

±16 

±1 

±8 

±j 

±4 

±4 

±2 

±1 

±1 

±2 

±4 

±3 

±1 

±6 

±i 

±8 

±  i 

±16 

±A 

Fio.  214. 

8.  Plot  the  curve  for  y=x\  and  thus  find  the  curve  from  which  cnn 
be  read  cubes  and  cube  roots  of  numbers. 

9.  In  simple  interest,  if  p  stands  for  principal,  /  for  time,  r  for  rate, 
and    a   for    amount,    then  a  =  p(l+rO.     If  now  particular  numerical 
values  are  given  to  p  and  r,  and  if  the  different  values  of  a  be  taken  as 
ordinates,  and  the  corresponding  values  of  t  as  abscissas,  then  the  graph 
of  this  equation  may  be  drawn.     Draw  the  graph.     What  line  in  the 
figure  represents  the  principal?     What  feature  in  the  graph  depends 
upon  the  rate  per  cent? 

10.  With  the  same  axes  as  used  in  exercise  9,  draw  the  graph  for  which 
interest  and  time  are  the  coordinates  of  points  on  the  curve. 


GRAPHICS 


363 


11.  Using  the  formula  for  the  area  of  a  circle  A  =irr2,  plot  the  curve, 
using  radii  as  abscissas  and  areas  as  oridnates,  from  which  can  be  read 
the  areas  of  circles  of  radii  from  0  to  6. 

12.  Follow  directions  similar  to  those  in  exercise  11  and  plot  the  curve 
from  which  can  be  read  the  volumes  of  spheres. 

13.  If  the  temperature  of  a  gas  is  constant,  the  pressure  times  the  vol- 
ume remains  constant.     Using  p  for  pressure  and  v  for  volume,  plot  so  as 
to  show  their  relative  change  when  pv  =  4. 

291.  Simpson's  Rule. — The  area  of  the  space  included 
between  a  curve  and  a  straight  line  can  easily  be  found  ap- 
proximately by  the  use  of  Simpson's  Rule  which  may  be 
stated  as  follows:  Let  AB,  in  Fig.  215,  be  the  curve  and  CD 
the  straight  line.  Divide  the  length  CD  into  an  even  num- 


m         o          r 


FIG.  215. 


ber  of  equal  parts,  say  8,  of  length  a,  and  erect  the  ordinates 
ho,hi,h2,  •  •  •  ,  hs.  Then  the  area  of  the  figure  CDBA  will 
be  given  by  the  formula: 


It  is  to  be  noticed  that  the  coefficients  of  the  ordinates  are 
alternately  4  and  2,  excepting  the  first  and  the  last.  The 
greater  the  number  of  divisions  made  the  more  accurate,  in 
general,  will  be  the  result. 

In  words  this  may  be  stated  in  the  following: 
RULE.  Divide  the  base  CD  into  an  even  number  of  equal 
parts,  and  measure  the  ordinate  at  each  point  of  division.  Add 
together  the  first  and  last  ordinates,  twice  the  sum  of  the  other  odd 
ordinates,  and  four  times  the  sum  of  the  even  ordinates;  multiply 
the  sum  by  one-third  the  distance  between  consecutive  ordinates. 
The  result  is  the  area  inclosed  (approximately). 


364 


PRACTICAL  MATHEMATICS 


292.  The  average  ordinate  rule. — For  approximate  results 
the  area  between  a  curve  and  a  straight  line,  base  line,  may 
be  found  as  follows: 

RULE.  Divide  the  base  line  into  any  number  of  equal  parts; 
at  the  center  of  each  of  these  parts  draw  ordinates.  Take  the 
average  length  of  these  ordinates  and  multiply  by  the  length  of 
the  base  line.  The  result  is  the  area  inclosed  (approximately). 

In  Fig.  215,  mn,  op,  rs,  etc.,  are  the  ordinates. 

A  convenient  way  for  adding  the  ordinates  is  to  draw  a  line 
of  indefinite  length;  then  with  the  dividers  measure  the 
ordinates  successively  on  this  line.  The  total  length  can  then 
be  measured  at  once.  This  will  avoid  errors  to  some  extent. 

293.  Area  in  a  closed  curve. — Either  of  the  methods  given 
may  be  used  in  finding  the  area  within  a  closed  curve.     Thus, 


FIG.  216. 

in  Fig.  216,  draw  the  two  parallel  tangents  OY  and  MN,  and 
draw  OX  perpendicular  to  these.  Divide  ON  into  any  number 
of  equal  parts1  (an  even  number  for  Simpson's  Rule)  and  draw 
the  ordinates  AB,  CD,  etc.  Call  the  several  widths  of  the 
figure  on  these  ordinates  hi,  hi,  As,  etc.  These  widths  can  be 
used  in  Simpson's  Rule  to  find  the  area  within  the  closed  curve. 

The  widths  mn,  op,  rs,  etc.,  can  be  used  in  the  average 
ordinate  rule  to  find  the  area. 

1  To  divide  a  line  into  any  number  of  equal  parts  see  Art.  149. 


GRAPHICS  365 

It  should  be  noted  that  h0  and  ^8  for  this  figure  are  each  0. 

294.  The  steam  indicator  diagram. — As  a  useful  application 
of  the  discussion  in  the  preceding  articles,  we  will  consider  the 
steam  indicator  diagram.1 

The  steam  indicator  is  a  mechanical  device  to  attach  to  a 
steam  engine  to  make  a  graphical  representation  of  the  steam 
pressure  acting  on  the  piston  throughout  the  stroke.  Knowing 


FIG.  217. 


the  pressure,  the  indicated  horse-power  of  the  engine  can  be 
calculated  from  the  formula  given  in  exercise  4,  page  313, 


rr 

H  =  33^000'  Where 

H  —  indicated  horse-power, 

P  =  mean  effective  pressure  in  pounds  per  square  inch, 
L  =  length  of  stroke  ;n  feet, 
A  =  area  of  piston  in  square  inches, 
N  =  the  number  of  strokes  per  minute. 
The  indicated  horse  -power  is  the  power  developed  by  the 
steam  on  the  piston  of  the  engine,  without  any  deduction  for 
friction. 

The  effective  horse-power  is  the  actual  available  horse- 
power delivered  to  the  belt  or  gearing,  and  is  always  less  than 
the  indicated  horse-power,  because  the  engine  itself  absorbs 
some  power  by  the  friction  of  its  moving  parts. 
The  indicator  diagram  may  be  as  given,  Fig.  217. 

1  For  a  full  discussion  of  the  steam  indicator  the  student  is  referred 
to  Peabody  —  Manual  of  the  Steam  Engine  Indicator. 


366 


PRACTICAL  MATHEMATICS 


The  width  of  a  rectangle  the  same  in  length  as  this,  and  of  the 
same  area,  would  represent  the  mean  effective  pressure  per 
square  inch  on  the  piston  during  the  stroke. 

The  diagram  is  always  to  a  certain  scale,  which  is  known 
from  the  indicator.  For  instance,  the  scale  might  be  60  Ib. 
per  inch  in  diagram.  The  mean  effective  pressure  is  then 
00  Ib.  multiplied  by  the  average  width  of  the  indicator  diagram. 

The  average  width  of  the  diagram  may  be  found  by  dividing 
the  area  by  the  length.  The  area  can  be  found  by  Simpson's 
Rule  or  by  the  average  ordinate  rule. 

A  convenient  method  for  locating  the  ordinates  is  to  place 
a  common  ruler  as  in  Fig.  218,  and  locate  10  ordinates,  the  two 


FIG.  218. 

end  ordinates  being  half  as  far  from  either  end  as  the  distance 
between  the  other  ordinates. 

The  average  length  of  these  ordinates  multiplied  by  the 
scale,  taken  from  the  indicator,  gives  the  mean  effective 
pressure. 

Example.  Taking  the  indicator  diagram  in  Fig.  218,  find 
the  mean  effective  pressure  of  the  steam  if  the  scale  is  30  Ib. 
to  the  inch.  Find  the  horse-power  of  the  engine  if  the  diam- 
eter of  the  piston  is  18  in.,  length  of  stroke  2\  ft.,  and  number  of 
revolutions  110  per  minute.  (The  number  of  strokes  of  the 
piston  is  twice  the  number  of  revolutions.) 

Solution.  Adding  together  the  ten  ordinates,  we  have 
1.82+2.66+2.81  +  2.91+2.73+2.21  +  1.78+1.46+1.17+0.70 
=  20.25. 


GRAPHICS  367 

Since  there  are  10  ordinates,  the  mean  is  20.25  -r-  10  =  2.025. 
Multiplying  by  the  scale,  we  have  2.025X30  =  60.75  =  pounds 
pressure  per  square  inch,  the  answer  to  the  first  part. 

Area  of  piston  =  3.1416X92  =  254.47  in.2 

t      ^  u     PLAN 

Formula  for  the  horse-power  is  H  = 


=  60.75,    L  =  2.5,     A  =  254.47, 
60.75X2.5X254.47X220 


EXERCISES  103 

1.  An  indicator  diagram  has  a  length  of  2  in.     The  ten  ordinates, 
beginning  at  the  left,  are  0.70  in.,  0.90  in.,  0.97  in.,  0.85  in.,  0.67  in., 
0.52  in.,  0.42  in.,  0.35  in.,  0.23  in.,  0.07  in.     Draw  a  diagram  that  these 
ordinates  will  satisfy.     If  the  indicator  scale  is  120  Ib.  to  the  inch,  find 
the  mean  effective  pressure.  Ans.  68.16  Ib.  per  in.2 

2.  Find  the  indicated  horse-power  of  an  engine  having  the  indicator 
diagram  of  exercise  1,  if  length  of  stroke  is  3  ft.,  diameter  of  piston  23  in., 
and  number  of  strokes  100  per  minute.  Ans.  257.4+  h.  p. 


FIG.  219. 

3.  Draw  a  semicircle  2  in.  in  radius.     Divide  the  diameter  into  four 
parts  and  find  the  area  by  Simpson's  Rule.     Divide  the  diameter  into 
ten  parts  and  find  area  by  the  same  rule.     Find  the  area  by  the  formula. 
Compare  the  three  results  and  state  your  conclusions. 

4.  Find  the  area  of  the  ellipse  in  Fig.  219  by  Simpson's  Rule.     Find 
the  area  by  the  formula  A  =irab,  and  compare  the  two  results. 

6.  Find  the  area  of  the  indicator  diagram  of  Fig.  217  both  by  Simpson's 
Rule  and  by  the  average  ordinate  rule,  using  ten  divisions.  Compare 
the  results.  Ans.  4.46  in.2  nearly. 

6.  If  the  indicator  diagram  of  exercise  5  has  a  length  of  3.73  in.,  and 
the  scale  is  50  Ib.,  find  the  mean  effective  pressure.  If  the  stroke  is 
2  ft.  6  in.,  the  cylinder  18  in.  in  diameter,  and  the  number  of  revolutions 
per  minute  80.  find  the  indicated  horse-power  of  the  engine. 

Ans.  59.79  Ib.  per  in.2;  184.4  +  h.  p. 


PART  FOUR 

LOGARITHMS  AND 
TRIGONOMETRY 


CHAPTE-R  XXXIV 
LOGARITHMS 

295.  Uses. — By  the  use  of  logarithms,  the  processes  of  mul- 
tiplication, division,  raising  to  powers,  and  extracting  roots 
of   arithmetical  numbers  are  much  simplified.     The  process 
of  multiplication  becomes  one  of  addition,  that  of  division 
becomes  one  of  subtraction,  that  of  raising  to  a  power  be- 
comes one  of  simple  multiplication,  and  that  of  extracting  a 
root  becomes  one  of  simple  division. 

Many  calculations  that  are  difficult  or  impossible  by  ordi- 
nary arithmetical  methods  are  readily  carried  out  by  means  of 
logarithms.  For  instance,  by  the  help  of  logarithms  a  square 
root  is  more  readily  found  than  by  ordinary  methods,  and  any 
other  root  is  found  as  easily  as  a  square  root.  The  value  of  a 

17 

number  affected  with  any  exponent,  as  2.34     ,  can  be  com- 
puted easily  by  logarithms. 

296.  Exponents. — For  convenience  the  definitions  and  laws 
of  exponents  previously  given  are  repeated  here. 

Definitions. 

(1)  an  =  a-a-a  •  •  -  to  n  factors,     n  an  integer. 

(2)  <•-»  =  -• 

(3)  o»  =  l. 

/    A   \  "  Wl  / 

(4)  a*  =  a-\/an. 

24  369 


370  PRACTICAL  MATHEMATICS 

Laws. 

(1)  an-am  =  a""H". 

(2)  an-s-am=an-">. 

(3)  (a-6-c-  •  •  )"  =  an6"cn-  •  •  • 

(A\     faV       a" 

(4)  (b)   "V 

(5)  (an)m  =anm. 

297.  Definitions  and  history. — A  logarithm  of  a  number  is 
the  exponent  by  which  the  base  must  be  affected  to  produce 
that  number. 

The  logarithms  of  all  the  positive  numbers  to  a  given  base 
are  called  a  system  of  logarithms,  and  the  base  is  called  the 
base  of  the  system. 

Any  base  may  be  used  in  a  system  of  logarithms;  but  the 
base  10  is  commonly  used  because,  as  will  be  seen  later,  it 
makes  a  very  convenient  system  of  logarithms  to  work  with. 

Logarithms  were  invented  by  John  Napier  of  Scotland,  who 
lived  from  1550  to  1617.  They  were  described  by  him  in  1614. 
Napier  used  the  number  2.71828  •  •  •  as  a  base.  This  base 
is  still  used  in  mathematics  (see  Art.  316) . 

Henry  Briggs  (1556  to  1631),  professor  at  Gresham  College, 
London,  modified  the  new  invention  by  using  the  base  10,  and 
so  made  it  more  convenient  for  practical  purposes. 

Because  of  the  time  they  save  and  the  help  they  give  in 
performing  difficult  computations,  logarithms  may  be  consid- 
ered among  the  great  inventions  of  the  world. 

298.  Notation. — If  we  take  2  as  a  base,  we  may  write  in  the 
language  of  exponents,  24  =  16.     In  the  language  of  logarithms, 
we  may  express  the  same  idea  by  saying,  the  logarithm  of  16 
to  the  base  2  is  4.     This  is  abbreviated  and  written  thus 
Iog2  16=4. 

Similarly,  we  have  the  following  expressed  in  the  language  of 
exponents  and  in  the  language  of  logarithms: 

Language  of  Language  of 

exponents  logarithms 

25  =     32.  log,       32  =  5. 

34=     81.  log,       81=4. 

54  =   625.  Iog5     625=4. 


LOGARITHMS  371 

Language  of  Language  of 

exponents  logarithms 

83  =   512.  logs     512  =  3. 

4°-5  =       2.  Iog4         2  =  0.5. 

8'=       4.  logs         4  =  §. 

64'  =       4.  Iog64        4  =  i 

103  =  1000.  Iog101000  =  3. 


EXERCISES  104 

Answer  as  many  as  you  can  orally.     Cover  the  answers  so  that  they 
will  not  be  seen  until  the  results  are  written. 

1.  Express  the  following  in  the  language  of  logarithms: 

(1)  26=64.  (4)   16°-5  =  4.  (7)  32°-4=4. 

(2)  53  =  125.  (5)   104  =  10,000.  (8)   10l-3979=25. 

(3)  73  =  343.  (6)   125' =5.  (9)   102-5465  =  352. 

Ans.   (1)  Iog264=6.        (4)  logie  4  =0.5.  (7)  Iog32  4=0.4. 

(2)  logs  125=3.      (5)  logio  10,000  =  4.         (8)  logio  25  =  1.3979. 

(3)  logy  343  =3.      (6)logi255  =  i  (9)  logio  352=2.5465. 

2.  Express  the  following  in  the  language  of  exponents: 

(1)  log,  256=8.  (4)  logio  643  =2.8082. 

(2)  loge  216=3.  (5)  Iog10  429  =  2.6325. 

(3)  logie     2=0.25.  (6)  logic  999  =2.9996. 
Ans.   (1)     2s      =256.  (4)  102-8082=643. 

(2)  63      =216.          (5)  102-6325=429. 

(3)  16°-25  =  2.  (6)  102-9996  =  999. 

3.  Find  the  logarithms  of  the  following : 

(1)  log;  49.  (4)  Iog9  729.  (7)  Iog8  2. 

(2)  logs  243.  (5)  Iog4  256.  (8)  Iog16  64. 

(3)  logs  3125.  (6)  log,0  100,000.  (9)  log,0  0.01. 

Ans.  (1)  2;  (2)  5;  (3)  5;  (4)  3;  (5)  4;  (6)  5;  (7)  *;  (8)  4;  (9)  -2. 

4.  Find  the  value  of  x  in  the  following : 

(1)  logs   x  =  4.  (5)  logic  x=  -2.  (9)  Iog25  z=i 

(2)  logio  z  =  6.  (6)  logs  x=  -3.  (10)  logie  z=f. 

(3)  logie  a;  =  i  (7)  logio  x=  -3.  (ll)logmz  =  i 

(4)  logio  z=0.  (8)  logs  x=  |.  (12)  Iog49  x  =  |. 

Ans.   (1)  81;  (2)  1,000,000;  (3)  8;  (4)  1;  (5)   0.01;   (6)  ^r;  (7)  0.001; 
(8)  16;  (9)  125;  (10)  64;  (11)  25;  (12)  343. 


372  PRACTICAL  MATHEMATICS 

6.  Find  the  value  of  x  in  the  following  : 

(1)  log,  100-2.  (5)  log,  4-0.5.  (9)  log,  27-0.75. 

(2)  log,  81  =4.  (6)  log,  4-0.25.  (10)  log,  2  =  0.125. 

(3)  log,  512  =  3.  (7)  log,  16  ~J.  (11)  log*  49  =  §. 

(4)  log,  1024  =  10.  (8)  log,  17-0.5.  (12)  log,  100  -J. 
Ana.  (1)  10;  (2)  3;  (3)  8;  (4)  2;  (5)  16;  (6)  256;  (7)  32;  (8)  289;  (9)  81; 

(10)  256;  (11)  343;  (12)  1000. 

299.  Illustrative  computations  by  means  of  exponents.  — 
It  is  very  important  that  the  fundamental  ideas  of  logarithms 
shall  be  well  understood.  In  this  article  are  given  examples 
illustrative  of  the  use  of  exponents,  or  logarithms,  in  making 
computations  in  multiplication,  division,  raising  to  powers, 
and  extracting  roots.  These  computations  are  made  by  the 
help  of  the  following  table  in  which  2  is  the  base.  The  work 
is  kept  in  the  language  of  exponents  but  can  be  easily  trans- 
lated into  the  language  of  logarithms. 

2-6  =  &  2!=2  2"  =256 

2-&  =  &  2*  =4  2»  =512 

2-'  =  ^  23  =  8  2'°  =  1024 

2~3  =  t  2<  =  16  2"  =  2048 

2-J  =  J  25  =  32  21*  =  4096 

2-'  =  J  2«  =  64  2"  =  8192 

2°    =1  27  =  128  2M  =  16,384 

Multiplication.     Multiply  512  by  32. 
From  the  table,  512  =  29, 
and32  =  26. 


From  the  table,   2"  =  16,384. 
.'.  512X32  =  16,384. 

Diinsion.     Divide  512  by  4096. 
From  the  table,  512  =  29, 
and  4096  =  212. 
.'.  512-^4096  =  2*  H-212  =  2-3. 
From  the  table,  2~3  =  J- 
.'.  512  -h  4096  =  I 

Raising  to  a  power.     Find  the  value  of  16s. 
From  the  table,    16  =  24. 

.'.  163  =  (2«)3  =  2". 
From  the  table,  212=4096. 

.'.  16s  =  4090. 


2.  256X64. 
5.  32X256. 
8.  2048  -=-256. 
11.  163. 

3.  2048X8. 
6.  &X  16,384. 
9.  16^-512. 
12.  642. 

14.  A/4096. 
16,384X512 

15.  \/  16,  384. 
32X64 

2048 
20.   (32X128)?- 

512X8 
21.  (8192X32)1 

LOGARITHMS  373 

Extraction  of  a  root.     Find  the  value  of  ^4096. 
From  the  table,  4096  =  212. 


From  the  table,       23=8. 
.'.  ^4096  =  8. 

EXERCISES  105 

By  the  help  of  the  table  on  page  372  find  the  value  of  the  following. 
Do  the  work  without  using  a  pencil  when  possible. 

1.  256X16. 

4.  128X128. 

7.  8192  -H  1024. 
10.  8-^512. 
13.  Vl^384. 

^j*?.0!? 
512 

19.   (4096)1 

1024X64  /  512X256  \i  3  1256X1024 

512X16,384'  '    \1024  X8192/  \      40962 

300.  Logarithms  of  any  number.  —  It  is  readily  seen  that,  in 
the  exercises  of  the  last  article,  the  numbers  considered  were 
all  integral  powers  of  2.     It  is  also  readily  seen  that  there  are 
many  numbers  that  cannot  be  expressed  as  integral  powers  of 
2.     The  same  thing  is  true  for  any  base.     Thus,  for  the  base  3, 
we  have  as  integral  powers  the  numbers  3,  9,  27,  81,  243,  etc., 
and  no  numbers  except  these  between  3  and  243. 

We  can,  therefore,  with  any  given  base  write  integral  log- 
arithms for  only  a  small  part  of  all  possible  numbers.  That 
is,  the  logarithms  of  numbers  to  any  given  base  are  usually 
not  integral.  Thus,  the  logarithm  of  95  to  the  base  3  is  4 
and  some  fraction,  because  95  is  between  34  and  35.  What  this 
fraction  is  cannot  easily  be  determined. 

301.  Logarithms  to  the  base  10.—  In  what  follows,  if  no 
base  is  stated,  it  is  understood  that  the  base  10  is  used. 

When  the  base  is  10  we  evidently  have  the  following: 
log  100,000  =  5.  log      1        =0. 

log     10,000  =  4.  log  0.1  -1. 

log        1000  =  3.  log  0.01      =-2. 

log          100  =  2.  log  0.001    =-3. 

log  10  =  1.  log  0.0001  =-4. 


374  PRACTICAL  MATHEMATICS 

The  logarithm  of  any  number  between  1000  and  10,000 
is  between  3  and  4,  or  it  is  3  and  some  fraction.  Between  100 
and  1000  the  logarithm  is  2  plus  a  fraction.  Between  0.01 
and  0.1  the  logarithm  is  —2  plus  a  fraction  or  —1  minus  a 
fraction.  In  order  that  the  fractional  part  of  the  logarithm 
may  always  be  positive,  we  shall  agree  to  take  the  logarithm  so 
that  the  integral  part  only  is  negative. 

In  general,  the  logarithm  of  a  number  consists  of  two  parts, 
a  whole  number  part  and  a  fractional  part. 

The  whole  number  part  is  called  the  characteristic. 

The  fractional  part  is  called  the  mantissa. 

The  mantissas  of  the  positive  numbers  arranged  in  order 
are  called  a  table  of  logarithms. 

The  logarithm  of  3467  consists  of  the  characteristic  3  and 
some  mantissa  because  3467  lies  between  1000  and  10,000. 
The  logarithm  of  59,436  is  4  plus  a  fraction  because  59,436 
lies  between  10,000  and  100,000.  The  log  0.0236  is  -2  plus 
a  fraction  because  0.0236  lies  between  0.01  and  0.1. 

It  is  readily  seen  that  multiplying  a  number  by  10  increases 
its  characteristic  by  1. 

One  of  the  great  advantages  in  using  the  base  10  is  that  the 
characteristics  can  be  determined  by  inspection.  It  is  only 
necessary,  then,  to  have  the  mantissas  given  in  a  table. 

302.  Rules  for  determining  the  characteristic. — From  what 
has  been  said  in  the  last  article,  and  from  a  further  con- 
sideration of  the  table  given  there,  the  following  rules  are 
evident: 

(1)  For  whole  numbers,  the  characteristic  is  one  lefts  than  the 
number  of  whole  number  figures  and  is  positive. 

(2)  For  decimals,  the  characteristic  is  one  more  than  the  num- 
ber of  zeros  immediately  at  the  right  of  the  decimal  point  and 
is  negative. 

(3)  In  a  number  consisting  of  a  whole  number  and  a  decimal, 
consider  the  whole  number  part  and  apply  rule  (1). 

Thus,  the  characteristics  of  the  following  are  as  given: 
of  326  is  2  by  rule  (1),  of  37,265  is  4  by  rule  (1),  of  0.046  is 
-2  by  rule  (2),  of  0.000046  is  -5  by  rule  (2),  of  2.36  is  0  by 
rule  (3),  and  of  276.36  is  2  by  rule  (3). 


LOGARITHMS  375 

EXERCISES  106 

State  the  characteristics  to  the  base  10  of  the  following: 

846  3956  2.325  87,654 

44.36  43,968  0.0123  9.3264 

173.94  39.267  0.00492  0.0003967 

4.7654  3333.3  0.4689  0.039643 

303.  The  mantissa. — The  determination  of  the  mantissa  is 
more  difficult  than  the  determination  of  the  characteristic. 
The  mantissa  is  found  from  a  table  of  logarithms. 

Tables  of  logarithms  are  made  only  by  a  great  deal  of  work. 
They  are  spoken  of  as  three  place  tables,  four  place  tables, 
ten  place  tables,  etc.,  according  to  the  number  of  decimal 
places  given  in  the  mantissas.  The  degree  of  accuracy  in 
computations  made  by  logarithms  depends  upon  the  number 
of  places  in  the  table  used,  the  more  places  in  the  table  the 
greater  the  degree  of  accuracy.  The  tables  generally  used 
are  those  having  from  four  to  six  places. 

The  mantissa  depends  only  upon  the  figures  of  the  number, 
and  not  at  all  upon  the  decimal  point.  To  illustrate  this, 
consider  the  following:  Suppose  that  we  have  given  that  log 
867  =  2.9380. 

(1)  This  means  that  102-9380  =  867. 

(2)  Now  102  =  100. 

(3)  Dividing  (1)  by  (2),  10°-9380  =  8.67. 

(4)  /.  log  8.67  =  0.9380. 

Hence  the  mantissa  of  8.67  is  the  same  as  the  mantissa  of 
867. 

It  is  evident  that  if  both  members  of  (1)  are  multiplied  or 
divided  by  any  integral  power  of  10  the  mantissa  is  unchanged. 
Hence  the  decimal  point  of  a  number  can  be  moved  as  we 
please  without  the  mantissa  of  the  logarithm  of  the  number 
being  changed. 

This  also  illustrates  the  fact  that  a  change  in  the  position 
of  the  decimal  point  of  a  number  changes  the  value  of  the 
characteristic  of  the  logarithm  of  the  number. 

304.  Tables. — Upon  examining  a  four  place  table  of  loga- 
rithms (see  Table  X),  it  is  noticed  that  the  first  column  has 
the  letter  N  at  the  top.     This  is  an  abbreviation  for  number. 
The  other  columns  have  at  their  tops  and  bottoms  the  numbers 


376  PRACTICAL  MATHEMATICS 

0,  1,  2,  3,  •  •  •  9.  Any  number  consisting  of  three  figures  has 
its  first  two  figures  in  the  column  headed  N  and  its  third 
figure  at  the  top  of  another  column.  For  instance,  take  the 
number  456;  45  is  found  in  the  column  headed  N  and  6  at  the 
top  of  another  column. 

The  columns,  after  the  first,  are  made  up  of  numbers  con- 
sisting of  four  figures.  These  numbers  are  decimals,  and  are 
the  mantissas  of  the  logarithms  of  the  numbers  made  up  of 
the  figures  in  the  column  headed  N  together  with  a  figure  from 
the  head  of  another  column. 

The  difference  between  two  consecutive  mantissas  is  called 
the  tabular  difference,  that  is,  the  table  difference. 

305.  To  find  the  mantissa  of  a  number. — 

(1)  When  the  number  consists  of  three  significant  figures. 
Example.     Find  the  mantissa  of  347. 

From  the  manner  in  which  the  table  is  formed,  the  first  two 
figures  of  347  are  found  in  the  column  headed  N,  and  the  third 
figure  at  the  top  of  the  page.  The  mantissa  of  347  is  found  to 
the  right  of  34  and  in  the  column  headed  7.  It  is  0.5403. 

The  mantissa  of  3.47,  3470,  or  any  number  consisting  of 
these  figures  in  the  same  order  is  0.5403. 

(2)  When  the  number  consists  of  one  or  two  significant  figures, 
the  number  is  found  in  the  column  headed  N,  and  the  mantissa 
to  the  right  in  the  column  headed  0. 

Thus,  the  mantissa  of  13  is  0.1139,  and  the  mantissa  of  4  is 
0.6021;  this  is  found  to  the  right  of  40. 

(3)  When  the  number  consists  of  four  or  more  significant 
figures. 

Example  1.     Find  the  mantissa  of  7586. 

Since  7586  lies  between  7580  and  7590,  its  mantissa  must  lie 
between  the  mantissas  of  7580  and  7590. 
Mantissa  of  7580  =  0.8797. 
Mantissa  of  7590  =  0.8802. 

The  difference  between  these  mantissas  is  0.0005,  which  is  the 
tabular  difference.  Since  an  increase  of  10  in  the  number 
increases  the  mantissa  C.0005,  an  increase  of  6  in  the  number 
will  increase  the  mantissa  0.6  as  much,  or  the  increase  is  0.0005 
X  0.6  =  0.0003.  Hence  the  mantissa  of  7586  =  0.8797+0.0003  = 
0.8800. 


LOGARITHMS  377 

The  process  of  determining  the  mantissa  as  above  is  called 
interpolation.  As  carried  out,  the  mantissa  is  supposed  to 
increase  in  a  constant  ratio  between  the  values  taken;  however, 
this  supposition  is  only  approximately  true. 

Example  2.     Find  the  mantissa  of  43,286. 
Mantissa  of  43,200  =  0.6355. 
Mantissa  of  43,300  =  0.6365, 
and  the  tabular  difference  =  0.0010. 

Since  an  increase  of  100  in  the  number  increases  the  mantissa 
0.0010,  an  increase  of  86  in  the  number  increases  the  mantissa 
0.0010X0.86  =  0.0009,  to  the  nearest  fourth  decimal  place. 
Hence  the  mantissa  of  43,286  =  0.6355+0.0009  =  0.6364. 

The  processes  that  have  been  given  should  seem  reasonable ; 
but,  since  the  process  of  finding  a  mantissa  has  to  be  performed 
so  often,  it  is  best  to  do  it  by  rule. 

306.  Rules  for  finding  the  mantissa. — 

(1)  For  a  number  consisting  of  three  figures,  find  the  first 
two  figures  of  the  number  in  the  column  headed  N,  and  the  third 
figure  at  the  head  of  a  column;  then  read  the  mantissa  in  the  column 
under  the  last  figure  and  at  the  right  of  the  first  two  figures. 

(2)  For  a  number  consisting  of  one  or  two  figures,  find  the 
number  in  the  column  headed  N,  and  the  mantissa  opposite  in 
the  column  headed  0. 

(3)  For  a  number  consisting  of  more  than  three  figures,  find 
the  mantissa  for  the  first  three  figures  by  rule  (1),  and  add  to  this 
the  product  of  the  tabular  difference  by  the  remaining  figures  of 
the  number  considered  as  decimals. 

Illustrations.  The  mantissa  of  243,  2.43,  0.0243,  or  any 
number  consisting  of  these  three  figures  in  the  same  order,  is 
found  by  rule  (1)  to  be  0.3856. 

The  mantissa  of  25,  0.025,  or  any  number  consisting  of 
these  two  figures  in  this  order,  is  found  by  rule  (2)  to  be  0.3979. 

The  mantissa  of  2364  is  found  by  rule  (3)  to  be  0.3736. 

Process.  Mantissa  of  2360  =  0.3729.  Tabular  difference  is 
0.0018.  0.0018  X  0.4  =  0.0007.  Adding  this  to  the  mantissa  of 
2360  gives  the  mantissa  of  2364  =0.3736. 

307.  Finding  the  logarithm  of  a  number. — In  looking  up 
the  logarithm  of  a  number,  it  is  best  to  first  determine  the 
characteristic,  and  then  the  mantissa. 


378  PRACTICAL  MATHEMATICS 

Example  1.     Find  the  logarithm  of  236. 
By  rule  (1)  for  characteristic  we  find  2. 
By  rule  (1)  for  mantissa  we  find  0.3729. 
.'.log  236  =  2.3729. 

Example  2.     Find  the  logarithm  of  7326. 
Rule  (1)  for  characteristic  gives  3. 
Rule  (3)  for  mantissa  gives  0.8649. 
/.log  7326  =  3.8649. 

Example  3.     Find  the  logarithm  of  0.00037. 
Rule  (2)  for  characteristic  gives  —4. 
Rule  (2)  for  mantissa  gives  0.5682. 
/.log  0.00037  =  4.5682. 

It  is  not  permissible  to  place  the  minus  sign  before  the 
characteristic  in  writing  a  negative  logarithm,  for  this  would 
indicate  that  both  characteristic  and  mantissa  are  negative, 
and  we  have  agreed  that  the  mantissa  shall  always  be  positive. 
To  overcome  the  difficulty,  the  negative  sign  is  placed  above  the 
characteristic.  Another  way  of  writing  the  negative  loga- 
rithm is  to  increase  the  characteristic  by  10  and  subtract  10 
at  the  right  of  the  mantissa. 

Thus,  the  logarithm  of  0.00037  is  written  4.5682  or 
6.5682-10. 

EXERCISES  107 

1.  Study  the  following  to  fix  in  mind  the  meaning  of  characteristic  and 
mantissa : 


log 

4580=3.6609; 

that  is, 

4580 

Id 

. 

6809 

log 

458.0  =  2.6609; 

that  is, 

458.0 

- 

in 

i, 

8809 

log 

45.80  =  1.6609; 

that  is, 

45.80 

- 

in 

i  . 

6809 

log 

4.580=0.6609; 

that  is, 

4.580 

10 

•  •• 

6809 

log 

0.4580=1.6609; 

that  is, 

0.4580 

= 

111 

T 

.8809 

log 

0.0458=2.6609; 

that  is, 

0.0458 

=  10Z 

.8809 

log 

000458=3.6609; 

that  is, 

0.00458 

= 

LO 

, 

.8609 

2.  Verify  the  following  by  the  tables: 

(I)  log  10  =  1.0000.  (2)  log  100=2.0000. 
(3)     log  110  =  2.0414.  (4)  log  2=0.3010. 
(5)     log  20  =  1.3010.  (6)  log  200  =  2.3010 
(7)     log  0.2  =  1.3010.  (8)  log  542  =2. 7340. 
(9)     log  345  =  2. 5378.  (10)  log  5.07-0.7050. 

(II)  log  78.5  =  1.8949.  (12)  log  0.981  =  1.9917. 


LOGARITHMS  379 

(13)  log  1054  =  3.0228.  (14)  log  1272=3.1045. 

(15)  log  0.0165  =2.2175.  (16)  log  0.1906  =  1.2801. 

(17)  log  21.09  =  1.3241.  (18)  log  0.09095=2.9588. 

(19)  log  3.060  =  0.4857.  (20)  log  4.411  =0.6445. 

(21)  log  07854=1.8951.  (22)  log  0.10125=1.0054. 

(23)  log  54.657  =  1.7377.  (24)  log  0.09885  =2.9950. 


308.  To  find  the  number  corresponding  to  a  logarithm. — 

In  nearly  every  problem  involving  logarithms  it  is  not  only 
necessary  to  find  the  logarithms  of  numbers,  but  the  inverse 
process,  that  of  finding  a  number  corresponding  to  a  logarithm, 
has  to  be  performed. 

Since  the  decimal  point  in  no  way  affects  the  mantissa,  we 
can  determine  only  the  figures  of  the  number  from  the  mantissa. 
The  decimal  point  has  to  be  placed  by  the  rules  for  determining 
the  characteristic. 

(1)  When  the  mantissa  of  the  given  logarithm  is  exactly  given 
in  the  table.     As  an  example,  find  the  number  having  2.8344 
for  a  logarithm. 

Find  in  the  table  the  mantissa  0.8344.  To  the  left  of  this 
mantissa,  in  the  column  headed  N,  find  the  first  two  figures, 
68,  of  the  number,  and  at  the  head  of  the  column  in  which  the 
mantissa  is  found,  find  the  third  figure,  3,  of  the  number.  The 
number  then  consists  of  the  figures  683,  but  we  do  not  know 
where  the  decimal  point  is  till  we  consider  the  characteristic. 

Since  the  characteristic  is  2,  there  must  be  three  figures  at 
the  left  of  the  decimal  point.  Hence  the  number  having 
2.8344  for  a  logarithm  is  683. 

This  means  that  102'8344  =683.  Notice  that  a  change  in 
the  characteristic  would  change  the  position  of  the  decimal 
point.  Thus,  the  number  corresponding  to  4.8344  is  68,300; 
while  the  number  corresponding  to  2.8344  is  0.0683. 

(2)  When  the  mantissa  of  the  given  logarithm  is  not  exactly 
given  in  the  table.     As  an  example,  find  the  number  corre- 
sponding to  the  logarithm  3.4689. 

Find  the  mantissas  0.4683  and  0.4698,  between  which  the 
given  mantissa  lies.  The  number  corresponding  to  3.4698  = 
2950.  The  number  corresponding  to  3.4683  =  2940.  That  is, 
an  increase  in  the  mantissa  of  0.0015  makes  an  increase  of  10 
in  the  corresponding  number.  The  given  mantissa  is  0.0006 


PRACTICAL  MATHEMATICS 

larger  than  0.4683.  Then  the  required  number  is  <j;oo?$ X10 
=  4  larger  than  2940.  Hence  the  number  corresponding  to 
the  logarithm  3.4689  is  2944. 

In  dealing  with  the  tabular  difference,  for  convenience,  it  is 
best  to  drop  the  decimal  point.  Then  we  should  have 
AX10  =  4  instead  of  818^X10  =  4. 

309.  Rules  for  finding  the  number  corresponding  to  a  given 
logarithm.— (1)  When  the  mantissa  of  the  given  logarithm  is 
exactly  given  in  the  table,  the  first  two  figures  of  the  number  are 
found  to  the  left  of  the  given  mantissa  in  the  column  headed  N, 
and  the  third  figure  is  found  at  the  head  of  the  column  in  which 
the  mantissa  is  given. 

(2)  When  the  mantissa  of  the  given  logarithm  is  not  exactly 
given  in  the  table,  find  the  mantissa  nearest  the  given  mantissa 
but  smaller.  The  first  three  figures  of  the  number  are  those 
corresponding  to  this  mantissa,  and  are  found  by  rule  (1). 

For  another  figure,  divide  the  difference  between  the  mantissa 
found  and  the  given  mantissa  by  the  tabular  difference.  The 
quotient  is  the  other  figure.  Always  determine  this  figure  to  the 
nearest  tenth. 

In  both  (1)  and  (2)  place  the  decimal  point  so  that  the  rules 
for  determining  the  characteristic  may  be  applied  and  give  the 
given  characteristic. 

Example  1.  Find  the  number  of  which  2.8414  is  the  log- 
arithm. The  mantissa  0.8414  is  found  in  the  table  to  the  right 
of  69  and  in  the  column  headed  4;  hence  the  number  consists 
of  the  figures  694.  The  decimal  point  must  be  placed  so  as  to 
give  a  characteristic  of  2  when  the  rule  for  characteristic  is 
applied.  Hence  694  is  the  number  whose  logarithm  is  2.8414. 

Example  2.  Find  the  number  whose  logarithm  is  1.7624. 
Mantissa  nearest  0.7624  is  0.7619  which  is  the  mantissa  of 
578.  Tabular  difference  =  8.  Difference  between  the  man- 
tissas is  5.  5-7-8  =  0.6  nearly.  Hence  1.7624  is  the  log  57.86. 

EXERCISES  108 

Find  the  value  of  x  or  verify  the  following: 
1.  0.3010=log*.  2.  1.6021  =log  x. 

3.  2 . 9031  =log  x.  4.  1 . 6669  =  log  46.44. 

6.  2.7971=log626.7.  6.  3 . 9545  =  log  9006. 

7.  0.8794=  log  7.575.  8.  3.9371  =  log  x. 


LOGARITHMS  381 

9.  0.8294=  log  6.  752.  10.  1  .9039=  log  80.15. 

11.  9.  3685  -10=  log  x.  12.  8.  9932  -10=  log  0.09845. 

13.  8.  9535  -10=  log  x.  14.  7.  7168-  10  =log  0.00521. 

15.  6.7016-10=log  0.000503.  16.  7  .  8654  -  log  x. 

17.  3.4792=log  0.003014.  18.  4.9231  =log  0.0008378. 

19.  4.  2975  =log  0.0001984.  20.  4.2975=log  z. 

310.  To  find  the  product  of  two  or  more  factors  by  the  use 
of  logarithms.  —  RULE.  Find  the  sum  of  the  logarithms  of  the 
factors.  The  product  is  the  number  corresponding  to  this  sum. 

Example.     Find  the  product  of  3.76  X  0.89  X  7.628. 
Process,     log  3.76    =0.5752 
log  0.89   =1.9494 
log  7.628  =  0.8824 
log  of  product  =  1.4070 
.'.  product  =  25.  53. 

That  this  is  an  application  of  the  law  of  exponents  in  mul- 
tiplication will  be  seen  from  the  following  form: 

3.76  =  10°'5752 

0.89  =  10T'9494 

7.628  =  10°'8824 

.'.  3.76X0.89X7.628=  100<5752X10T-9494X10 

_  -i  rvO.  5  7  5  2+T.  9494+0-8824 


311.  To  find  the  quotient  of  two  numbers  by  logarithms.— 

RULE.  Subtract  the  logarithm  of  the  divisor  from  the  logarithm 
of  the  dividend.  The  quotient  is  the  number  corresponding  to 
this  difference. 

Example  1.     Find  the  quotient  of  38.76  -f-  7.923. 
Process,     log  38.76   =1.5884 
log     7.923  =  0.8989 
log  of  quotient  =  0.6895 
.',  quotient  =  4.892. 


382  PRACTICAL  MATHEMATICS 

7.246X0.8964X5.463 
Example  2.     Evaluate  -j^—^-^— 

Process. 

log  7.246   =0.8601  log    4.27     =0.6304 

log  0.8964=9.9525-10    log    0.3987  =  9.6007-10 
log  5.463   =0.7374  log  27.89     =1.4454 

log  of  Num.  =  1.5500  log  of  Den.  =  1.6765 

log  of  Den.  =  1.6765 

log  of  quotient      =1.8735 
.'.  quotient  =0.7473. 


EXERCISES  109 

1.  Multiply  the  following  by  the  use  of  logarithms  : 

(1)  226X85  =  19,210.  (2)  7.25X240  =  1740. 

(3)  3272X75  =  245,400.  (4)  0.892X805  =  718.1. 

(5)   1.414X2.829=3.999.  (6)  42.37X0.235  =  9.958. 

(7)  2912X0.7281=2120.  (8)  289X0.7854  =  227. 

(9)  7.62X3.67=27.97.  (10)  7.09X3.99=28.29. 

(11)   10.00124X89.5=0.1110.  (12)  4.768X9.872=47.07. 

2.  Divide  by  the  use  of  logarithms  and  check  by  actual   division: 
(1)  3025+55.  (2)  0.2601-7-0.68. 

(3)  3950  +  0.250.  (4)   10  +  3.14. 

(5)  0.6911+0.7854.  (6)  1+762. 

(7)  6786  +  4236.  (8)  200  +  0.5236. 

(9)300  +  17.32.  (10)0.220+0.3183 

3.  Find  the  product  of  3.246X98.768X0.4376.  Ana.  140.3. 

4.  Find  the  product  of  0.00389X9.876X0.00468.      Ans.  0.0001798. 
6.  Divide  the  product  of  0.38765  and  7.498  by  4.3792.    Ans.  0.6637. 

Evaluate  the  following  by  the  use  of  logarithms: 

110X3.1X0.650  . 

6-  33X0.7854X1.7 
6000X5X29 
r<  0.7854X25,000X81.7 

3.516X485X65^ 
J<  3.33X17X18X73 
0  15X0.37X26.16  ft  , 

9-  11X8X18X6^7'  An8'  °-01374- 

in     78  X  52X1605 
10"  338X767X431- 


,    u 
An9' 


11        05X3.15X428  A 

"*  0.317X0.973X43.7 


LOGARITHMS  383 

312.  To  find  the  power  of  a  number  by  logarithms. — RULE. 
Multiply  the  logarithm  of  the   number  by  the   exponent  of  the 
power.     The  number  corresponding  to    this    logarithm  is    the 
required  power. 

Example  1.     Find  the  value  of  (2.378) G. 
.Process,     log  2.378  =  0.3762 

GXlog  2.378  =  2.2572  =  log  of  the  power. 

.'.  (2.378)  6  =  180.8. 

Example  2.     Find  value  of  (9.876) }. 
Process,     log  9.876  =  0.9946 

f  of  log  9.876  =  0.7460  =  log(9.876)f. 

.'.    (9.876)f  =  5.571. 

313.  To  find  the  root  of  a  number  by  logarithms. — RULE. 

Divide  the  logarithm  of  the  number  by  the  index  of  the  root.     The 
number  corresponding  to  this  logarithm  is  the  required  root. 

Example  1.     Find  ^27.658. 
Process,     log  27.658  =  1.4418 

i  log  27.658  =  0.2884  =  log  v/27.658- 
.'.  ^27.658=1.943. 

Example  2.     Find  v/0.008673. 
Process,     log    0.008673  =  7.9382-10 

log  v/0.008673  =  |  of  (7.9382-10) 
=  |  of  (57.9382-60) 

=  9.6564  -10. 

.*.  ^0.008673  =  0.4533. 

Note  that,  as  in  this  example,  when  we  are  to  divide  a  loga- 
rithm with  a  negative  characteristic,  not  a  multiple  of  the 
divisor,  it  is  best  to  first  add  and  subtract  such  a  number  of 
times  10,  that  after  dividing  there  will  be  a  — 10  at  the  right. 
Thus,  in  the  above,  before  dividing  (7.9382  —  10)  by  6,  we  add 
and  subtract  50. 

314.  Computations  made  by  logarithms  only  approximate.— 
By  performing  the  above  operations  without  the  use  of  loga- 
rithms, it  will  be  noticed  that  the  results,  in  most  cases,  will 
vary  slightly  from  those  obtained  by  the  use  of  logarithms. 
This  emphasizes  the  fact  that  computations  by  the  use  of 


384  PRACTICAL  MATHEMATICS 

logarithms  are  only  approximately  correct.  However,  as 
great  a  degree  of  accuracy  as  desired  can  be  obtained  by  using 
tables  of  a  large  enough  number  of  places.  In  general,  we 
obtain  an  answer  with  about  as  many  correct  figures  as  the 
number  of  places  in  the  logarithm  tables. 

315.  Natural  logarithms. — While  in  computations  it  is 
usually  more  convenient  to  employ  logarithms  to  the  base  10, 
often  in  theoretical  work,  and  in  some  formulas,  logarithms 
to  another  base  are  used.  This  base  is  a  number  that  cannot 
be  exactly  expressed  by  figures.  To  seven  decimal  places  it 
is  2.7182818.  It  is  usually  represented  by  the  letter  e,  just 
as  the  ratio  of  the  circumference  of  a  circle  to  its  diameter  is 
represented  by  the  Greek  letter  v.  Logarithms  to  this  base 
are  called  natural  logarithms,  hyperbolic  logarithms,  or 
Napierian  logarithms. 

Tables  of  natural  logarithms  are  published;  but,  for  pur- 
poses of  computation,  it  is  only  necessary  to  remember  that 
the  natural  logarithm  of  a  number  is  approximately  2.3026 
times  the  common  logarithm  of  the  same  number.  Or  the 
common  logarithm  of  a  number  is  0.4343  times  the  natural 
logarithm.  Stated  in  symbols,  this  is,  in  the  exponential 
form: 

10n  =  2.718282'3026n,  or  2.71828"  =  10° 4343n; 
and  in  logarithmic  form  it  is 

\ogcN  =  2.3026  log10AT,  or  log,oAT  =  0.4343  log,.V, 

where  N  is  any  number. 

As  an  example  of  the  occurrence  of  the  Napierian  logarithms 
in  a  problem  of  applied  mathematics,  consider  the  following: 
In  finding  the  insulation  resistance  by  the  leakage  method,  the 
resistance  R  is  found  from  the  formula 


"  i       (V>\ 
log.(T°) 


where  t  is  time  in  seconds,  C  capacity,  and  V  voltage. 

Compute  the  value  of  R  if  <  =  120,    F0=123.    V  =  115.8, 
and  C  =  0.082. 


LOGARITHMS 


385 


Solution.     In  order  that  we  may  use  the  table  of  common 
logarithms,  the  formula  may  be  written 

_     e  t  l 

C 


2.3026 


Substituting  values, 

ff-1Q.X     12°    • 

A  .082  ' 


1 


log  115.8) 


2.3026(log  123 
.-.  R  =  2.426  X1010. 

As  other  examples  where  the  base  of  the  natural  system  of 
logarithms  occurs,  consider  the  following:  (1)  In  an  alter- 
nating current  circuit,  the  current  i  at  any  instant  is  given 
by  the  formula 


where  /  is  the  maximum  current,  R  the  resistance,  L  the  coeffi- 
cient of  self-induction,  t  the  time  in  seconds,  and  e  the  base 
of  the  natural  system  of  logarithms. 

(2)  The  work  W  done  by  a  volume  of  gas,  expanding  at  a 
constant  temperature,  from  a  volume  F0  to  a  volume  Vi  is 
given  by  the  following  formula: 


Formulas  that  involve  logarithms  to  the  base  e  occur  fre- 
quently in  applications  of  calculus. 

EXERCISES  110 


Use  logarithms  in  evaluating  the  following  : 


1. 
2. 


(0.543)3. 
(4.07)3. 

3.  (1.738)3. 

4.  (1.02)5. 
6.  GfWx)3. 

6.  Uj)4. 

7.  (0.1181)*. 

8.  (1381)*. 

9.  (1024)  A. 
18.  (7.23) 7-23. 

19. 


Ans.  0.1601. 

Ans.  67.42. 

Ans.  5.248. 

Ans.  1.104. 

Ans.  0.004394. 

Ans.  2.868  X lO"10. 

Ans.  0.3436. 

Ans.  11.14. 

Ans.  128. 


10.  (1.4641)*. 
(0.00032)*. 

-v/2. 


11. 
12. 

13.  A/4. 

14.  A/3. 
16.  A/2^ 

16.  A/0.03 

17.  (4.56)4 


20. 


48X64X11 

522X300 

12  X. 31225X400,000' 
25 


Ans.  1.1. 

Ans.  0.2. 
Ans.  1.260. 
Ans.  1.414. 
Ans.  1.442. 
Ans.  1.149. 
Ans.  0.5023. 
Ans.  1012 

Ans.  1,627,000. 
Ans.  0.2899. 

Ans.  0.5411. 


PRACTICAL  MA  THE  MA  TICK 


21     J *°° 

\55X3.1416* 


22.  50  X; 


23.   x/ -431X96', 
\64X1500 


24. 


(1.06)4 
3.8961  X. 6945  X. 01382 


Ans.  1.622. 
Ana.  4.769. 

.  19.60. 

».  51.14. 


4694  X. 00457 
26.  -C/. 0009657  -^70044784". 

/  7.61  X. 0593  U 
'    \       1.307       /' 

28.  v/5i06.5  x  .dobosHoaT. 

29     /^°_V 
'    W37J 

30.  (837.5 X. 0094325)'. 

31.  (.01)* -H  ^7. 
.000561 6X^/424^65 


Ans.  0.001743. 
Ans.  1.070. 
Ans.  0.4505. 
.  0.7945. 
5.259. 


32. 


33. 


34. 


(6.73)<X(.03194)i 
/3929X>^6548 
-\X72L83 


.05287 


Ans.  1.805. 
Ana.  0.0005228. 

Ans.  0.00001146. 
Ans.  188.2. 
An*.  1.023. 


v'TSTTXv'. 078359 

P<  __  T  D 

35.  In  the  formula  S  =  ^~>  find  R  if  5  =  500,  £  =  220,  and  7  =  12. 


.'2-20 


Ans.  9.17. 


Suggestion.  Here,  as  in  many  exercises  which  follow,  the  computation 
can  be  more  easily  done  without  logarithms  than  with.  Endeavor  to 
use  logarithms  only  when  they  are  necessary  or  when  they  save  time. 

36.  Using    the    formula    for    the    horse-power  of  a    steam   engine, 

PLAN    . 
H=  33000'  (^  Art.  261): 

(1)  Find  H  when  P  =  76.5,  L  =  2},  A  =231.8,  and  #  =  116. 

(2)  Find  N  when  H  =52,  P  =49.12,  L  =  1.5,  and  A  =  113.1. 

Ans.  140.3;  205.9. 

37.  Given  TF  =  .0033XlO-7n,  find  H'  when  n  =  75,000. 

Ans.  0.00002475. 

38.  If  £  =  E.  M.  F.  in  volts  in  moving  conductor,  L=  length  of  con- 
ductor in  centimeters,   V  =  velocity  in  centimeters  per  second,  B=  the 
number  of  lines  of  force  per  square  centimeter,  we  have  the  formula 

E^LVBIQ-*.     Given    V'--^-.  £-8000,  and  L -0.6X100,  find  £. 

Ans.  0.072. 


60 


LOGARITHMS  387 

39.  Find  the  value  of  P  from  the  following  formula  when  A  =  11, 

~  A  Kf 
=  1.71,  and  C  =  1.3:          p  =  3/— ^    _—  Ans.  0.6199. 


Suggestion.     First   find    the   logarithm    of   the  expression   under  the 
radical,  and  then  divide  by  3.     This  will  give  the  log  P. 

40.  In  measuring  electrical  resistance  by  a  Wheatstone's  bridge,  the 
following  data  were  taken  : 

(1)  R  =  8,  d  =  539.7,  a2=459. 

(2)  R  =  Q,  ^  =510.1,  a2  =488.4. 

Find  the  values  of  x  from  (1)  and  (2)  by  the  formula 

!)  9.404;  (2)  9.40. 

41.  In  finding  the  diameter  of  a  wrought-iron  shaft  that  will  transmit 
90  horse-power  when  the  number  of  revolutions  is  100  per  minute,  using 
a  factor  of  safety  of  8,  we  have  to  find  the  diameter  d  from  the  formula 

90~ 
=  68.5X     /  —   —  ennnn'     Find  the  value  of  d.  Ans.  3.591. 


Wgl3 

42.  Find  the  value  of  M  from  the  formula  M  =  .,  ,3p>  when  gr  =  980, 

TF  =  75,  Z  =  50,  6=0.98178,  d  =  0.5680,  and  5  =  0.01093. 

Ans.   11.69X10". 
Solution. 

log  W        =   1.8751  log  4         =0.6021 

log  g          =  2.9912  log  b         =1.9920 

log  Z3          =   5.0970  logd3        =1.2629 

log  Num.  =  9.9633  log  B        =2.0386 

log  Den.    =   3.8956  log  Den.   =3.8956 

log  M         =  12.0677 
.'.  M         =1,169,000,000,000  =  11.  69X10". 

43.  Find  the  value  of  n  from  the  formula  n  =  -  2    4   ,  when  g  =  980, 

1  =  28,  9  =  0.857,  r  =  0.477,  L  =  109.7,  and  m  =  100.       Ans.  2.476X10". 

44.  Use  the  same  formula  as  in  exercise  43,  and  find  the  value  of  n 
when  L  =  69.6,  m  =  10,  gr  =  980,  Z  =  28,  0  =  1.1955,  and  r  =0.317. 

Ans.  0.577X10". 

45.  If  m  =  ar-1-16,  find  r  when  TO  =2.263  and  a  =  0.4086. 

Ans.  0.2287. 
Solution.     Solving  for  r,  we  have 


log  a  =1.6113 
log  m=  0.3547 

log -=1.2566 


388 


PRACTICAL  MA  THE  MA  TICS 


It  is  now  best  to  unite  the  characteristic  and  mantissa  by  algebraic 
addition. 

Thus,  1.2566= -1 +.2566  =-.7434. 

Now  multiply  by  the  exponent  .862,  and  get  -. 7434 X. 862=  -.6408. 

This  is  next  changed  to  the  form  of  a  logarithm  as  given  in  the  table, 
that  is,  to  a  form  where  the  mantissa  is  not  negative. 
-.6408=1.3592. 

.MJ    .  _ 

=  1.3592. 


=  0.2287.     Ana. 


46.  Given  p  =  pc 


(2   \  — 
—_^-t  )  y~  1,  find  the  value  of  p  in  terms  of  puif  -y  : 


\V+IJ 


1.41. 


FIG.  220. 


An*.  p=0.527p0. 

(For  the  meaning  of  this 
formula  see  Perry's  Calculus, 
page  55.) 

47.  If  an  open  tank  kept  full 
of  water  has  a  rectangular  notch 
cut  on  one  side  as  shown  in  Fig. 
220,  the  number  of  cubic  feet  of 
water  that  will  flow  through 
*n*s  notch  P61"  second  is  given 
by  the  formula: 


where      Q  =  the  amount  of  flow  in  cubic  feet, 

c  =  a  constant  found  by  experiment, 

6  =  the  width  of  the  notch  in  feet, 

h  =the  depth  of  the  notch  in  feet, 
and  0  =  32.2. 

Find  Q  when  A  =  1J,  b  =  2,  and  c=  0.586.  Am.  9.654. 

48.  Using  the  formula  of  the  last  exercise,  find  Q  when  h  =  I,  b  =  2.5, 
and  c  =  0.589.  Aiis.  5.991. 

49.  If  Q  is  the  number  of  cubic  feet  that  will  flow  through  a  V-shaped 
notch  per  second  and  h  the  height  in  feet  of  the  water  above  the  bottom 
of  the  notch,  then  Q  «  hi     If  Q  =  7.26  when  h  =  1  .5,  find  h  when  Q  =  5.68. 

Ana.  1.360. 

Suggestion.     If  Q  «  A*  then  Q  =  Kh*. 
Substituting  values,  7.26  =  A'1.5I.     /.  A'  =  7.26  -f-  1.5*. 
LTsing  this  value  of  K  with  Q  =5.68,  gives 

5.68  =  (7.26  -=-1.5')^. 
,    _5.68X1.5» 
7.126 

50.  The  following  is  an  approximate  formula  for  determining  the  num- 
ber of  wires  that  can  be  enclosed  in  a  pipe  : 


N  -0.907  /^-0.94\  '+3.7, 


LOGARITHMS  389 


where      N  =  the  number  of  wires, 

D  =  the  diameter  of  the  enclosing  pipe, 
d  =  the  diameter  of  the  wires. 


Solve  this  formula  for  D  and  find  D  =d  [  0.94  + 


0.907  . 

51.  Use  the  formula  of  the  preceding  exercise  and  find  the  diameter 
of  a  casing  to  hold  100  wires  each  having  a  diameter  of  i  in. 

Ans.  1.405  in. 

52.  The  amount  of  a  principal  at  compound  interest  for  a  certain  time 
is  given  by  the  following  formula: 

A=P(l+rY, 

where  A=the  amount,  P  =  the  principal,  r  =  the  rate  per  cent,  and 
t  =the  time  in  years. 

Find  the  amount  of  $236  at  compound  interest  for  14  years  at  3  per 
cent.  Ans.  $356.50. 

63.  Find  the  amount  of  $3764  at  compound  interest  for  21  years  at 
4J  per  cent.  Ans.  $9478. 

Remark.  It  should  be  noted  that  in  such  problems  as  those  of  exer- 
cises 52,  53,  and  the  following,  a  four-place  table  of  logarithms  will  not 
give  results  that  can  be  relied  upon  when  the  exponent  is  large.  For 
instance,  if  exercise  52  be  computed  by  a  six-place  table  of  logarithms 
the  amount  is  $356.97,  and  that  of  exercise  53  is  $9486.07.  Of  course, 
all  that  is  necessary  to  secure  a  desired  degree  of  accuracy  is  to  use  a  table 
of  logarithms  that  has  a  sufficiently  large  number  of  decimal  places.  In 
life  insurance'  computations  a  ten-place  table  is  often  used. 

54.  The  government  Farm  Loan  Banks  loan  money  to  farmers  for  a 
period  of  years  at  5  or  6  per  cent,  and  the  money  is  paid  back  in 
equal  annual  installments.  The  following  table,  in  use  by  the  United 
States  government,  gives  the  annual  installment  necessary  to  discharge 
a  loan  of  $1000  with  interest,  the  number  of  years  and  the  rate  of  interest 
being  indicated  in  the  table. 

Table  fcr  Loan  of  $1000 


Number  of 

years 
10  

Annual  ir 

at  5% 
$129.50 

tstallment 
at  6% 
$135.87 
102.96 
87.18 
78.23 
72.65 
68.97 
66.46 

15 

96.34 

20         

80  .  24 

25  

70  .  95 

30 

65  .  05 

35  

61.07 

40.. 

58.28 

Compute  the  installments  in  this  table  by  using  the  following  formula: 

Pr(l+r)» 


390  PRACTICAL  MATHEMATICS 

where  P~  the  amount  of  the  loan, 

r-=the  rate  per  cent  per  annum, 
n  =  the  number  of  annual  payments, 
;>  =  lh(>  amount  of  each  installment. 

Suggestion.     If  P  =  $1000,  r  =  6  %,  and  n  =  10,  then 
D     1000X.06X1.0610 
P=  -M36.87. 


65.  The  above  formula  will  give  the  amount  of  each  installment  made 
at  any  equal  intervals  of  time,  in  order  to  discharge  a  debt,  if  r  divided  by 
the  number  of  intervals  per  year  is  used  for  the  rate.     Thus,  if  the  install- 

ments are  monthly,  jo  is  the  rate  per  month. 

Find  the  amount  of  the  equal  monthly  payments  to  discharge  a  debt 
of  $3000  in  5  years  if  the  rate  is  6%.  Ans.  $58.34. 

06 
Suggestion.     Here  P=  3000,  n  =  60,  and  rate  per  month  is  -.,,  =  0.005. 

\£i 

66.  Find  the  amount  of  the  equal  semi-annual  installments  to  discharge 
a  debt  of  $4500  in  7  years  at  6%.  Ans.  $399.30. 

67.  If  an  indebtedness  is  paid  in  installments,  the  payments  being 
equal  and  each  including  the  interest  to  the  date  of  the  installment,  then 
the  number  of  installments  necessary  to  pay  the  debt  is  given  by  solving 
the  formula  of  exercise  54  for  n  and  finding 

_logp-log(p-Pr) 

log(l-j-r) 
where  P  =  the  total  indebtedness, 

p  =  the  amount  of  one  installment, 
and  r  =  the  rate  per  cent  for  the  period  between  installments. 

How  many  semi-annual  installments  of  $600  each  will  it  take  to  dis- 
charge a  debt  of  $7000  bearing  5%  interest.  Ans.  14. 

Suggestion.     Here   P  =  7000,   p  =  600,  and  r=  0.025.     Substituting  in 
formula, 

log  600  -log  (600  -175) 

log  1.025 

_  2.7782  -2.6284    0.1498  _  ,  ,     . 
~~     "0.0107" 


68.  An  indebtedness  of  $1500  is  to  be  paid  in  installments  of  $15  p*>r 
month,  each  installment  to  cover  the  interest  to  that  date.     Find  the 
number  of  installments  if  the  interest  is  at  8  per  cent  per  annum. 

Ans.  164.5. 

69.  An  indebtedness  of  $1800  is  paid  in  installments  of  $5.00  per  week, 
each  payment  to  cover  the  accrued  interest  to  that  date.     How  many 
payments  will  be  required  if  the  interest  is  at  5  per  cent  per  annum,  and 
52  weeks  are  counted  as  one  year?  Ans.   461.5. 

60.  Find  the  radius  of  a  sphere  that  contains  5263  cu.  ft. 

Ans.  10.79  ft. 


LOGARITHMS 


391 


Suggestion.     Take  the  formula  for  the  volume  of  a  sphere,  V  —  -Jar3, 
and  solve  it  for  r. 


This  gives  r  =  -C/— — 


61.  In  the  equation  y=log  x,  find  the  values  of  y  corresponding  to 
the  values  0.5,  1,  2,  5,  10,  25,  50,  and  100  of  x.     Choose  a  pair  of  coordi- 
nate axes  and  plot  these  pairs  of  points.     This  will  give  a  curve  that  shows 
the  relation  between  a  number  and  its  logarithm.     (It  will  be  best  to 
choose  a  unit  on  the  y-axis  5  or  10  times  the  length.  of  the  unit  on  the 
z-axis.) 

62.  If  the  mixture  in  a  gas  engine  expands  without  gain  or  loss  of  heat, 
it  is  found  that  the  law  of  expansion  is  given  by  the  equation  pv1-37  =  C. 
Given  that  p  =  188.2  when  v  =  ll,  find  the  value  of  C,  then  plot  the  curve 
of  the  equation  using  this  value  of  C.     This  curve  shows  the  pressure 
at  any  volume  as  the  gas  expands.     Consider  values  of  v  from  11  to  23. 

Solution.     Given    pv1-37  =  C. 

Substituting    p  =  188.2  and  w  =  ll,  gives  C  =  188.2  Xll1-37. 

Computing  by  logarithms, 

log   188.2  =  2.2747 
1.37  log  11  =  1.4267 
log  C  =  3.7014 
C  =  5028. 

The  formula  is  then  pz;1-37  = 
5028 


Now  choose,  say,  six  values  of  v,  as  given  in  the  table,  from  11  to  23 
and  compute  the  corresponding  values  of  p. 


V  

11 

13 

15 

17 

20 

23 

p.  . 

188.2 

149.7 

123.1 

103.7 

82.98 

68.53 

These  values  are  plotted  in  Fig.  221  and  a  smooth  curve  is  drawn 
through  the  points.  From  this  curve  any  value  of  p  corresponding  to 
values  of  v  from  11  to  23  can  be  read.  Such  curves  when  accurately 
plotted  are  of  great  value  in  engineering. 

63.  D  is  the  diameter  of  a  wrought-iron  shaft  to  transmit  an  indicated 

horse-power  H  at  N  revolutions  per  minute.     Given  D=  \~~\r  >  plot  a 

curve  showing  the  relation  between  D  and  H,  from  H  =  10  to  H  =  80, 
when  N  is  100  revolutions  per  minute.  From  your  curve  find  the  diam- 
eters for  horse-powers  35,  47,  and  72. 

Remark.  Many  of  the  following  exercises  can  be  worked  without  the 
use  of  logarithms.  Some  will  review  ideas  previously  discussed.  Use 
logarithms  only  where  they  are  more  convenient. 


392 


PRACTICAL  MATHEMATICS 


64.  The  tractive  power  of  a  locomotive  is  found  by  the  formula: 
T     DPtL 

w  ' 

where      D  =  the  diameter  of  the  cylinder  in  inches, 

P=the  mean  pressure  of  the  steam  in  the  cylinder  in  pound* 

per  square  inch, 

L  =  the  length  of  the  piston  stroke  in  inches, 
W  =  the  diameter  of  the  driving-wheel  in  inches, 

and          T  =  the  tractive  force  upon  the  rails  in  pounds. 


r 

200 
150 
100 

CO 
40 
30 
20 
10 

V 

\ 

\ 

\ 

\ 

^ 

\ 

\ 

\ 

' 

\ 

5 

\ 

. 

N 

\ 

^ 

\ 

N 

^ 

X 

\ 

V 

- 

0 

1    2    3    4    5     C    78    9   10  11  12  13  14  15  16  17  18  19  20  a  22  23  24  25  26  27 
Values    of    v 

Fio.  221. 

Find  the  tractive  force  for  the  following  data: 

(1)  Z)  =  16in.,  P  =  901b.,  L  =  45in.,  JT  =  78in.; 

(2)  Z)  =  20  in.,  P  =  100  lb.,  L=54in.,  T7  =  84  in. 

Arts.  (1)  74,770;  (2)  128,570. 

66.  To  find  the  weight  which  a  column  of  soft  steel,  with  square  bear- 
ings, will  support  per  square  inch  of  cross  section  we  use  the  following 
formula: 

P  =  ~~(i2LF 
^SoOOOr'' 


LOGARITHMS  393 

where  P  is  the  weight  in  pounds;  L,  the  length  in  feet;  and  r,  the  radius 

L 

of  gyration  in  inches.     Find  P  if  -  =5.  Ans.  40,910  Ib. 

r 

45000  45000  45000 

Suggestion.     P  = 


1   i    (12L)'  144     /L\  *  144X5^ 

"T" oijrw-w-i_9       •*•  "T" ocrvnn    I  „  /  ' 


36000r2         '  36000  \r/  36000 

d 

66.  In  a  column  which  is  solid  and  circular  in  cross  section.  r  =  —i 

4 

where  d  is  the  diameter  of  the  cross  section  in  inches.  Find  the  weight 
such  a  column  of  soft  steel  will  support  per  square  inch  of  cross  section, 
if  L  =  16  ft.  and  d  =  6  in.  Find  the  total  weight  this  column  will  support. 

Ans.  30,920  Ib.  per  sq.  in.  ;  874,400  Ib. 

•\/d2-\-di2 

67.  In  a  hollow  cylindrical  column.  r  =  —         —  '  where  d  and  di  are 

4 

the  outer  and  inner  diameters  of  the  column  respectively  in  inches. 
Find  the  weight  the  column  in  the  preceding  exercise  will  support  per 
square  inch  of  cross  section  and  the  total,  if  it  is  hollow  and  the  shell  is 
1  in.  thick.  Ans.  34,220  Ib.  per  sq.  in.  ;  537,500  Ib. 

68.  For  a  column  whose  cross  section  is  a  regular  hexagon,  r  =  0.264d, 
where  d  is  the  diameter  of  the  circle  inscribed  in  the  hexagon.     Find  the 
weight  a  solid  hexagonal  pillar  of  soft  steel  and  square  bearings  will 
support,  if  the  length  is  12  ft.  and  the  edge  of  the  base  is  2  in. 

Ans.  276,900  Ib. 

Note.  The  safe  load  for  the  above  columns  is  one-fourth  or  one-fifth 
of  the  value  given.  The  formula  used  is  Gordon's  formula.  For  medium 
steel  the  45,000  of  the  formula  is  changed  to  50,000. 

69.  In  any  class  of  turbine,  let  P  be  the  power  of  the  waterfall;  H,  the 
height  of  the  fall;  and  n.  the  rate  of  revolution.     It  is  known  that  for  a 
particular  class  of  turbines  of  all  sizes,  n  <x.Hl-Z5P~°-s.     In  the  list  of  a 
certain  maker,  when  the  fall  equals  6  ft.,  and  horse-power  100,  the  num- 
ber of  revolutions  per  minute  is  50.     Calculate  n  for  a  fall  of  20  ft.  and 
75  horse-power.  Ans.  n  =  260  nearly  . 

70.  in  transmitting  power  by  means  of  a  belt  and  pulley,  if  the  belt 
embraces  180°  of  the  pulley,  the  number  of  horse-power  transmitted  is 
given  by  the  formula: 


33000 
where      H  =  horse-power, 

t  —  tension  in  pounds  per  inch  width  of  belt, 
w  =  width  of  belt  in  inches, 
and  s  =  speed  of  belt  in  feet  per  minute. 

Solve  for  values  of  t,  w,  and  s  in  terms  of  the  remaining  letters. 

33000//  .         33000#  .        33000// 

Ans.    t=  -  •  —  >  w  =  —       —i  s  =  —      — 
ws  is  tw 

71.  Using  the  formula  of  the  preceding  exercise,  find  t  when  H  =  50, 
w  =  10  in,,  and  s  =4000  ft.  per  minute.  Ans.  41.25  Ib. 


394  PRACTICAL  MATHEMATICS 

72.  To  determine  the  elevation  of  the  outer  rail  in  a  curve  in  a  railroad 
track,  the  following  formula  is  used: 


where       e  =  the  elevation  in  feet, 

G  =  gage  of  the  track  in  feet, 

F=*  velocity  of  train  in  feet  per  second, 
and          72  =  radius  of  curvature  of  the  curve  in  feet. 

Find  e  for  the  following  data,  if  G  is  4  ft.  8i  in.,  the  standard  gage 

(1)  72  =  5730  ft.,  (a)  7  =  20  mi.  per  hr.,  (b)   F  =  50  mi.  per  hr. 

(2)  72  =  2865  ft.,  (a)  V  =  15  mi.  per  hr.,  (b)   F=40mi.  per  hr. 

(3)  72  =  716.8  ft.,  (a)   F  =  25  mi.  per  hr.,  (b)   7  =  50  mi.  per  hr. 

Ans.  (1)  (a)  0.02ft.,  (b)  0.14ft. 

(2)  (a)  0.02ft.,  (b)  0.18ft. 

(3)  (a)  0.27  ft.,  (b)  1.10  ft. 

73.  In  long  water  pipes,  when  the  diameter  and  length  of  the  pipes 
are  constant,  that  is,  do  not  change,  the  amount  of  discharge  varies  as 
the  square  root  of  the  head.     How  many  times  must  the  head  H  be  in- 
creased to  double  the  amount  of  discharge  G?     To  make  the  discharge 
five  times  as  much?  Ans.  4;  25. 

Definition.     The  head  is  the  distance  the  source  of  supply  is  above  the 
point  of  discharge. 

74.  If  the  pipe  is  of  such  length  and  diameter  that  the  discharge  is  20 
gallons  per  minute,  what  will  it  be  if  the  head  is  doubled? 

Ans.  28.28+  gallons. 

75.  If  when  the  head  is  10  ft.,  the  discharge  through  a  certain  pipe 
is  50  gallons  per  minute,  what  must  be  the  head  so  that  the  same  pipe 
may  discharge  210  barrels  per  hour?  Ans.  48.  l>2  ft. 

76.  In  long  water  pipes,  when  the  lengths  of  the  pipes  are  the  same 
and  when  the  head  does  not  change,  the  amount  of  discharge  varies 
directly  as  the  square  root  of  the  fifth  power  of  the  diameter.     Using  D 
and  d  for  the  diameters,  and  G  and  g  for  the  amounts  of  discharge,  write 
this  relation  in  the  form  of  a  proportion.      Ans.  G  :0  =  \/7J*  :\/d*. 

77.  Write  the  above  relation  in  the  variation  form  both  with  and 
without  the  constant.     What  does  the  constant  include  in  it?     Would 
there  be  a  different  constant  for  each  length  of  pipe?     For  each  head? 

Ans.  G  =  K\/D*',  G  a^/D*. 

78.  If  the  length  and  the  head  remain  constant,  what  change  in  the 
discharge  will  be  caused  by  a  change  in  diameter  from  3  in.  to  4  in.? 

Ana.  2.05  times,  nearly. 
Suggestion.     Increased  in  the  ratio  -\/3*  :\/4*. 

79.  A  3-in.  pipe  100  ft.  in  length  with  a  certain  head  discharges  110 
gallons  per  minute;  find  the  discharge  from  a  5-in.  pipe  of  the  same  length 
and  head.  Ans.  394.5  gal.  per  minute. 

Suggestion.     1  1  0  :  x  »  \/3*  •'  \/5». 

80.  How  many  1-in.  pipes  will  it  take  to  discharge  the  same  amount 


LOGARITHMS  395 


as  one  6-in.  pipe?     Here  we  are  considering  long  pipes  so  use  N= 

where  N  =  the  number  of  small  pipes,  and  D  and  d  are  the  diameters  of 
the  large  and  the  small  pipes  respectively.     (See  exercise  9,  page  157.) 

Ans,  88.2. 

81.  In  long  water  pipes,  when  the  discharge  and  the  length  are  con- 
stant, the  head  will  be  inversely  as  the  fifth  power  of  the  diameter.     Us- 
ing H  and  h  for  heads,  and  D  and  d  for  the  diameters  respectively,  write 
this  relation  in  the  proportion  form.  Ans.  H  :h=db  :Z)6. 

82.  With  a  head  of  4.  1  ft.  and  a  length  of  100  ft.  a  3-in.  pipe  will  dis- 
charge 95.4  gallons  per  minute;  find  the  head  so  that  a  2-in.  pipe  of  the 
same  length  will  discharge  an  equal  amount.  Ans.  31.13  ft. 

83.  Using  H  and  D  for  head  and  diameter  respectively,  write  the  rela- 

tion given  in  exercise  81  in  the  variation  form.     Using  this  formula  solve 

jf 

exercise  82.  Ans.  H  =  fz' 

84.  In  long  water  pipes  when  the  head  and  the  diameter  are  constant 
the  discharge  will  be  inversely  as  the  square  roots  of  the  lengths.     Using 
G  for  discharge  and  L  for  length,  state  this  in  the  variation  form.     In 

the  proportion  form.  „       K      ~  /-       ,— 

Ans.  G  =  ~7'>  G'-a^Vl  :\/L. 


86.  In  a  long  water  pipe  of  certain  diameter  and  head,  the  discharge 
is  2000  gallons  per  minute.  How  many  gallons  will  be  discharged  per 
minute  under  the  same  head  and  the  same  size  of  pipe  if  the  length  is 
doubled?  If  six  times  as  long?  Ans.  1414  gal.;  816.5  gal. 

86.  In  long  water  pipes,  when  the  discharge  and  the  diameter  are 
constant,  the  head  varies  directly  as  the  length.     Using  the  letters  given 
in  the  preceding  exercises,  state  this  in  the  form  of  a  proportion,  and  in 
the  variation  form.  Ans.  H:h=L:l;  H  —  KL. 

87.  The  square  of  the  initial  velocity  of  a  projectile  in  feet  per  second 
varies  as  the  number  of  pounds  of  powder  in  the  charge  and  inversely  as 
the  weight  of  the  projectile.     If  5  Ib.  of  a  certain  kind  of  powder  will 
give  a  projectile  weighing  10  Ib.  an  initial  velocity  of  1850  ft.  per  second, 
how  great  a  velocity  will  50  Ib.  of  powder  give  an  80-lb.  projectile? 

Ans.  2068  ft.  per  second. 

Solution.  Let  p—  number  of  pounds  of  powder  in  charge,  w=  weight 
of  projectile,  and  v  =  velocity  in  feet  per  second. 

Then  t'2  =  fc-£. 

w 
When  p  =  5  and  w  =  10,  v  =  1850. 


=  fcjj,  and  &  = 
When  p  =  50  and  w  =  80, 


88.  Using  the  same  quality  of  powder  as  in  the  last,  find  the  charge 
necessary  to  give  a  1200-lb.  projectile  an  initial  velocity  of  2100  ft.  per 
second.  Ans.  773  Ib. 


TRIGONOMETRY 
CHAPTER  XXXV 
INTRODUCTION,  ANGLES 

316.  Introductory. — Each    advance    step    in    mathematics 
is  an  attempt  to  do  something  more  easily  than  it  could  have 
been  done  before,  or  to  accomplish  something  that  was  before 
impossible.     We  have  seen  that  many  problems  could   be 
worked  more  easily  by  algebra  than  by  arithmetic,  and  that 
many  other  problems  could  be  solved  by  algebra  that  could  not 
be  solved  by  methods  of  arithmetic. 

It  was  found  that  the  area  of  a  segment  of  a  circle  could  not 
be  obtained  by  geometry  except  in  a  few  special  cases;  by 
methods  of  trigonometry,  this  area  can  be  found  in  all  cases 
where  there  are  sufficient  facts  to  do  it  by  any  means.  By 
geometry,  one  side  of  a  right  triangle  can  be  found  if  the  other 
two  sides  are  known;  but  there  is  no  way  by  geometry  of 
finding  the  acute  angles  when  only  the  sides  are  known.  By 
trigonometry,  the  angles  as  well  as  the  sides  can  be  found. 
Many  such  illustrations  could  be  given  in  which  trigonometry 
is  a  more  powerful  tool  than  either  algebra  or  geometry. 
Trigonometry  is  based  upon  geometry,  but  makes  use  of  the 
methods  and  machinery  of  algebra. 

While  trigonometry  can  be  applied  at  once  to  the  solution 
of  various  practical  problems,  it  is  also  of  great  assistance  in 
other  branches  of  mathematics.  In  the  following  chapters 
will  l>e  given  some  of  the  direct  applications  of  the  subject. 

317.  Angles. — The  definition  of  an  angle  as  given  in  Art. 
90  admits  of  a  clear  conception  of  small  positive  angles  only. 
In  trigonometry  we  wish  to  deal  with  negative  as  well  as  posi- 
tive angles,  and  these  of  any  size  whatever.     We,  therefore, 
need  a  more  comprehensive  definition  of  an  angle. 

If  a  line  is  turned  about  a  fixed  point  in  the  line  and  kept  in 
the  same  plane,  it  is  said  to  generate  or  sweep  out  an  angle. 

396 


INTRODUCTION,  ANGLES 


397 


The  hand  of  a  clock  may  be  thought  of  as  the  line  that  is 
revolving  and  generating  the  angle. 

The  size  of  the  angle  is  determined  by  the  amount  of  turning 
made  by  the  line. 

If  the  line  turns  in  a  counter-clockwise  direction,  that  is, 
opposite  in  direction  to  the  hands  of  a  clock,  the  angle  described 


Initial  Side  y 


Initial  Side 


FIG.  222. 


is  called  a  positive  angle.  If  the  line  turns  in  a  clockwise 
direction,  the  angle  described  is  called  a  negative  angle. 

The  position  of  the  line  at  the  start  is  called  the  initial 
line  or  side,  and  the  final  position  is  called  the  terminal  line 
or  side. 

A  circular  arrow  drawn  between  the  two  lines  and  having 
its  head  in  the  terminal  line  shows  the  direction  of  turning  and 
the  size  of  the  angle. 


00 


FIG.  223. 

In  Fig.  222 (a),  the  line  OX  is  imagined  pinned  at  O  and 
turning  in  a  counter-clockwise  direction  to  the  position  OP. 
The  angle  described  is  positive,  and  is  read  angle  XOP, 
Notice  that  the  initial  line  is  read  first. 

In  Fig.  222(6),  the  line  OX  is  thought  of  as  turning  in  a 
clockwise  direction,  and  so  describing  the  negative  angle  XOP. 


PRACTICAL  MATHEMATICS 


D 


It  is  evident  that  the  idea  of  an  angle  given  in  this  article 
allows  it  to  be  of  any  value  whatever,  positive  or  negative. 

Thus,  an  angle  of  467°  is  one  complete  turn  and  107°.  It  is  shown 
in  Fig.  223(a). 

An  angle  of  — 229°  is  a  turn  of  229°  in  the  negative,  clockwise,  direc- 
tion. It  is  shown  in  Fig.  223(6). 

An  angle  of  720°  is  two  complete  turns  of  the  initial  line.  An  angle 
of  3760°  is  ten  complete  turns  and  160°  more. 

318.  Location    of    angles,    quadrants. — For     convenience 
in  locating  the  angles,  the  agreement  is  made  as  for  plotting 

(see  Art.  286).  Two  lines, 
X'X  &nd  F'FofFig.  224,  are 
drawn  at  right  angles  to  each 
other.  The  directions  of  the 
lines  and  the  location  of  the 
quadrants  are  as  in  the 
article  referred  to. 

If  the  positive  direction  of 
the  x-axis  is  taken  as  the 
initial  side,  the  angle  is  said 
to  be  in  the  first  quadrant  if 
its  terminal  side  lies  between 
OX  and  OF.  It  does  not 

matter  how  many  turns  are  made.     Thus  the  angles  of  40°, 
400°,  760°  all  lie  in  the  first  quadrant. 

Similarly,  if  the  terminal  side  lies  between  OF  and  OX', 
the  angle  is  said  to  be  in  the  second  quadrant.  If  the  terminal 
side  lies  between  OX'  and  OF',  the  angle  is  said  to  be  in  the 
third  quadrant.  If  the  terminal  side  lies  between  OF'  and 
OX,  the  angle  is  said  to  be  in  the  fourth  quadrant. 

Thus,  angle  XOA  is  in  the  first  quadrant. 

Angle  XOB  is  in  the  second  quadrant. 
Angle  XOC  is  in  the  third  quadrant. 
Angle  XOD  is  in  the  fourth  quadrant. 

If  the  terminal  side  falls  on  OX,  OY,  OX',  or  OY'  the  angle 
is  said  to  lie  between  two  quadrants. 

319.  Measurement  of  angles. — In  Chap.  XJJI,  the  three 
units  for  measuring  angles  are  discussed.     These  units  are  the 
right  angle,  the  degree,  and  the  radian. 


INTRODUCTION,  ANGLES  399 

By  definition,  the  radian  is  an  angle  of  such  size  that  when 
placed  at  the  center  of  a  circle  its  sides  intercept  an  arc  equal 
to  the  radius.  It  is  found  in  the  chapter  referred  to  that 
2-n-  radians  are  measured  by  a  whole 
circumference.  This  is  illustrated 
in  Fig.  225.  From  the  definition 
it  is  evident  that  the  size  of  the 
radian  does  not  depend  on  the  size 
of  the  circle. 

As  before  1  radian  =  57.29578°- 
=  57°  17'  44.8",  and  1°  =  0.017453 
+  radians. 

The  measurement  of  an  angle  by 
the  radian  unit  is  often  called  circular  measure  or  r-measure. 

7T 

Since  2r  radians  =  360°,  -n   radians  ==  180°,  ~  radians  =  90°, 

7T 

~  radians  =  60°,  etc.,  it  is  often  convenient  to  represent  some 

o 

of  the  most  frequently  used  angles  by  means  of  TT. 

In  using  circular  measure,  the  word  radian  is  usually  omitted. 

7T     7T 

Thus,  we  write  TT,  ~'  T»  3,  0.5,  meaning  in  each  case  so  many 

radians. 

To  convert  radians  to  degrees,  multiply  the  number  of  radians  by 

1  on 

-or  57.29578-. 

7T 

To  convert  degrees  to  radians,  multiply  the  number  of  degrees 
by  ~  or  0.017453  + . 

Example  1.     Reduce  2.5  radians  to  degrees,  minutes,  and 
seconds. 

Solution.     1  radian  =  57.29578°, 

/.  2.5  radians  =  2.5X57.29578°  =  143.2394°. 

To  find  the  number  of  minutes,  multiply  the  decimal  part 
3f  the  number  of  degrees  by  60, 

.'.  0.2394°  =  60X0.2394  =  14.364'. 
Likewise,  0.364'  =  60 X0.364  =  21.8", 

.'.  2,5  radians  =  143°  14' 22". 


400  PRACTICAL  MATHEMATICS 

Example  2.     Reduce  22°  36'  30"  to  radians. 

Solution.     First,  change  to  degrees  and  decimals  of  degrees, 

22°  36' 30"  =  22.6083°+. 
1°  =  0.01 7453+ radians, 
/.  22.6083°+ =22.6083X0.017453  =  0.3946-  radians. 

320.  Relations    between   angle,    arc,    and  radius. — From 
the  definition  of  the  radian,  it  is  evident  that  the  number  of 
radians  in  an  angle  at  the  center  of  a  circle  can  be  found  by 
dividing  the  length  of  the  arc  its  sides  intercept  by  the  length 
of  the  radius. 

arc 

That  is,  number  of  radians  in  angle  =  —     '- — 

radius 

In  Fig.  226,  angle  AOB  (in  radians)  =—  -. — pr-r- 

radius  OA 

If  6  (the  Greek  letter  theta)  stands  for  the  number  of  radians 
in  an  angle,  r  for  the  length  of  the  radius  of  a  circle,  having 
its  center  at  the  vertex  of  the  angle,  and  s  for  the  length  of  the 
arc  between  the  sides  of  the  angle;  then, 

Solving,  first  for  s  and  then  for  r, 

r=s  +  d. 

These  relations  are  important  as 
they  may  be  used  in  solving  many 
practical  problems. 

Example.  The  diameter  of  a  gradu- 
ated circumference  is  10  ft.,  and  the 

graduations  are  5  minutes  of  arc  apart;  find  the  distance, 
length  of  arc,  between  the  graduations  in  fractions  of  an  inch 
to  three  decimal  places. 

Solution.     By  formula,  s  =  rO. 
From  tho  example,  r  =  12X5  =  60in., 
and  B  =  0.01745XV%  =  0.00145+. 

Substituting  in  the  formula,  s  =  60X0.00145+  =0.087+. 
.'.  length  of  5'  arc  is  0.087+  in. 

321.  Railroad  curves. — In  the  United  States  it  is  customary 
to  express  the  curvature  of  railroad  tracks  in  degrees.     The 


INTRODUCTION,  ANGLES  401 

degree  of  a  curve  is  determined  by  the  central  angle  which  is 
subtended  by  a  chord  of  100  ft.  Thus,  in  a  circle,  a  5-degree 
curve  is  one  in  which  a  100-ft.  chord  subtends  a  central  angle 
of  5  degrees. 

In  curves  commonly  used,  the  error  is  slight  if  the  arc  is 
taken  in  place  of  the  chord.  Then,  assuming  that  1  radian  = 
57.30°,  the  radius  of  a  1-degree  curve  is  found  by  the  formula 
for  r  of  Art.  320.  Thus, 


57.  oO 

Hence  1  degree  of  curvature  gives  a  radius  of  5730  ft.  It 
follows  that  5730  divided  by  the  number  of  degrees  in  the  curve 
gives  the  radius  of  the  curve  ;  and  5730  divided  by  the  number 
of  feet  in  the  radius  gives  the  number  of  degrees  in  the  curve. 

EXERCISES  111 

1.  In  what  quadrant  is  each  of  the  following  angles:  27°,  436°,  236°, 
4372°,  -46°,  -324°,  -90°,  -476°,  -2342°,  |TT,  3x,  |TT,  4.3,  78.5,  82.3? 
Draw  each  angle. 

2.  Draw  the  terminal  side  of  237°,  and  give  value  of  a  negative  angle 
having  the  same  terminal  side.     How  many  such  angles  are  there? 

3.  Draw  the  following  angles:  76°,  25.6°,  425°,  5263°,   -25°,   -236°, 
-146°,  -935°. 

4.  Can  a  positive  and  negative  angle  each  have  its  terminal  side  in  the 
same  position?     Illustrate. 

5.  Express  the  following  angles  as  some  number  of  times  x  radians: 
30°,  45°,  54°,  60°,  81°,  90°,  120°,  135°,  150°,  180°,  210°,  225°,  240°,  270°, 
300°,  315°,  330°,  360°,  540°,  720°. 

.         7TTr3j7r7r97r:r27r3x5ir  7x     STT     4x     Sir     Sir     7ir 

6'   4'   10'   3'   20'   2'   "3"'   T'    "6~'   *"'    IF'   T'   ~3'   ~2'   ~3'   T' 

lllr    O        Q        A 

~7r~,  Zir,  OTT,  4ir. 

6.  Reduce  the  following  to  radians: 

(a)  47°,  (f)   135°, 

(b)  75°  30',  (g)   120°, 

(c)  16°  43'  10",  (h)   175°  45'  40", 

(d)  125°  46'  30",  (i)    95°  10'  10", 

(e)  62°  40',  (j)    127°  41'  50". 

Ans.  (a)   .820  +  ;   (b)   1.318  -;  (c)  .292-;  (d)  2.195  +  ;   (e)   1.094  -; 
(f)  2.356  +  ;  (g)  2.094  +  ;  (h)  3.068-;  (i)  1.661  +  ;  (j)  2.229-. 

26 


402  PRACTICAL  MATHEMATICS 

7.  Reduce  the  following  (1)  to  degrees  and  decimals  of  degrees  to  four 
places,  (2)  to  degrees,  minutes,  and  seconds: 

(a)  J»,  (d)  4.23,  (g)  0.125, 

(b)  fr,  (e)  2.76,  (h)  2.236, 

(c)  ST,  (f)  UT,  (i)  3.14159. 

An?.,    (a)  45°.  (d)  242°  21'  40".  (g)  7°9'  43". 

(b)  135°.  (e)   158°  8'  11".  (h)  128°  6'  48". 

(c)  150°.  (f)   191°  15"  (i)   180°. 

8.  How  many  radians  are  in  each  of  the  angles  of  a  right  triangle,  if 
one  of  the  acute  angles  is  36°  47'?  Ana.  \v;  .642-;  .929-. 

9.  How  many  degrees  in  each  of  the  angles  of  an  isosceles  triangle,  if 
the  angle  at  the  vertex  is  \ir  radians?  Ana.  30°,  75°. 

10.  Two  of  the  angles  of  a  triangle  are  respectively  f  and  §  of  a  radian. 
Find  the  number  of  radians  and  degrees  in  the  third  angle. 

Ans.  2.0749  radians  =  118°  52'  59". 

11.  If  an  angle  of  126°  at  the  center  has  an  arc  of  226  ft.,  find  the 
radius  of  the  circle. 

Solution.     Use  the  formula  r  =8  +  0. 
0  =  126X0.017453=2.199. 
r  =  226X2.  199  =  102.77. 
.'.radius  is  102.77  ft. 

12.  A  flywheel  20  ft.  in  diameter  has  an  angular  velocity  of  3r  per 
second.     Find  the  rim  velocity.  Ans.  94.25  ft.  per  sec. 

13.  The  circumferential  speed  generally  advised  by  makers  of  emery 
wheels  is  5500  ft.  per  minute.     Find  the  angular  velocity  per  second  in 
radians  for  a  10-in.  wheel.  Ans.  220  radians  per  sec. 

Suggestion.      Use  the  formula  0  =  s-t-r. 

14.  Solve  similar  problems  for  the  velocities  of  the  following: 

(a)  Ohio  grindstones,  advised  speed  2500  ft.  per  minute. 

(b)  Huron  grindstones,  advised  speed  3500  ft.  per  minute. 

(c)  Wood,  leather  covered,  polishing  wheels,  7000  ft.  per  minute. 

(d)  Walrus  hide  polishing  wheels,  8000  ft.  per  minute. 

(e)  Rag  wheels,  7000  ft.  per  minute. 

(f)  Hair  brush  wheels,  12,000  ft.  per  minute. 

15.  A  flywheel  of  4-ft.   radius  is  revolving  counter-clockwise  with  a 
circumferential  velocity  of  75  ft.  per  second.     Find  the  angular  velocity 
in  radians  per  second.  Ans.  18}   radians. 

Solution. 


16.  A  train  is  traveling  on  a  curve  of  half  a  mile  radius  at  the  rate  of 
30  miles  per  hour.  Through  what  angle  does  it  go  in  15  seconds?  Ex- 
press the  answer  in  both  radians  and  degrees. 

Ana.  0.25  radian  =  14°  19'  26". 

Suggestion.     Use  the  formula  6  =s  -i-r,  where  s  =  i  mile  and  r  =  J  mile. 


INTRODUCTION,  ANGLES  403 

17.  Find  the  radius  of  a  circle  in  which  an  arc  of  20  ft.  measures  an 
angle  of  2.3  radians  at  the  center.     In  this  circle,  find  the  angle  at  the 
center  measured  by  an  arc  of  3  ft.  8  in. 

Ans.  8.70-  ft.;  0.421+  radians. 

18.  Find  the  angular  velocity  per  minute  of  the  minute  hand  of  a  watch. 
Express  in  degrees  and  radians.  Ans.  6°  =0.1047+  radians. 

19.  A  train  of  cars  is  going  at  the  rate  of  15  miles  per  hour  on  a  curve 
of  600-ft.  radius;  find  its  angular  velocity  in  radians  per  minute. 

Ans.  2.2  radians. 

20.  A  flywheel  22  ft.  in  diameter  is  revolving  with  an  angular  velocity 
of  9  radians  per  second.     Find  the  rate  per  minute  a  point  on  the  cir- 
cumference is  traveling.  Ans.  5940  ft.  per  min. 

21.  Find  the  length  of  arc  which,  at  the  distance  of  1  mile,  will  subtend 
an  angle  of  10'  at  the  eye.     An  angle  of  1".     Ans.  15.36  ft. ;  0.0256  ft. 

22.  The  radius  of  the  earth's  orbit,  which  is  about  92,700,000  miles, 
subtends  at  the  star  Sirius  an  angle  of  about  0.4".     Find  the  approximate 
distance  of  Sirius  from  the  earth.  Ans.  478  X 1011  miles. 

23.  What  radius  has  a  5-degree  curve  in  a  railroad  track?     A  curve  of 
3°  15'?     If  the  radius  of  curvature  is  4550  ft.,  what  is  the  degree  of  curve? 

Ans.  1146  ft.;  1763+  ft.;  H°  nearly. 


CHAPTER  XXXVI 
TRIGONOMETRIC  FUNCTIONS 

322.  Sine,  cosine,   and   tangent    of  an  acute   angle. — In 
geometry  we  learn  of  certain  relations  between  the  sides  of  a 
triangle;  here  we  find  that  the  angles  are  related  to  the  sides 
in  a  certain  way.     These  relations  are  very  useful  in  the  con- 

A    struction    of    angles    and    in 
S^.        solving  various  triangles  and 
other  figures. 

De^ruYt'ons.  If  an  acute 
angle  XOA,  Fig.  227,  is 
taken,  and  from  P,  any  point 

X  in  OA,  a  perpendicular  QP  is 

^  drawn  to  OX,  a  right  triangle 

QOP  is  formed.     It  is  evident 

because  of  similar  triangles  (see  Art.  104)  that  the  following 
ratios  will  not  change  no  matter  where  the  perpendicular 
QP  may  be  drawn,  so  long  as  the  angle  XOA,  or  6,  does  not 
change. 

QP 
The  ratio  ^p  is  called  the  sine  of  angle  6,  written  sin  6. 

The  ratio  ~p  is  called  the  cosine  of  angle  8,  written  cos  6. 

QP 
The  ratio  ~~  is  called  the  tangent  of  angle  6,  written  tan  6. 

These  lines  can  be  measured,  say,  in  inches,  and  numerical 
values  of  these  ratios  can  then  be  found. 

323.  Ratios  for  any  angles. — The  same  ratios  may  be  written 
for  an  angle  in  any  quadrant.     Thus,  in  Fig.  228,  angle  XOA 
is  in  the  second  quadrant.     QP  is  the  perpendicular  drawn 
from  any  point  in  the  terminal  side  to  the  z-axis.     The  ratios 
are 

sin  XOA  =  -jtTp*  cos  XOA  =  pcp>  tan  XOA  =  ^« 
404 


TRIGONOMETRIC  FUNCTIONS 


405 


As  an  exercise  the  student  may  write  these  ratios  for  the 
angles  XOB  and  XOC. 

As  will  be  seen  in  the  next  article,  some  of  the  ratios  are 
positive  and  some  are  negative  numbers. 


324.  General  form  for  ratios. — Let  the  distance  from  0 
along  the  terminal  side  to  the  point  chosen  be  called  the 
distance,  represented  by  r,  and  always  considered  positive. 
The  length  of  the  perpendicular  to  the  z-axis  is  the  ordinate 


IVs 


FIG 


of  the  point,  positive  if  extending  above  and  negative  if  below 
the  x-axis,  and  is  represented  by  y.  The  distance  from  O 
to  the  foot  of  the  perpendicular  is  the  abscissa  of  the  point, 
positive  if  extending  to  the  right  of  the  origin  and  negative 


tor. 


PRACTICAL  MATHEMATKs 


if  to  the  left,  and  is  represented  by  x.     The  ratios  for  any 
angle  0  (see  Fig.  229)  may  then  be  written  as  follows: 

ordinate    y 
sin  6=  —         —  =  -» 
distance     r 

abscissa    x 
cos  6=  ..—         =    » 
distance    r 

ordinate     y 
tan  6  =   , 

abscissa    x 

The  reciprocals  of  these  ratios  are  often  used  and  are  named 
as  follows: 


cosecant  6,  abbreviated  esc  6  = 
secant  6,  abbreviated  sec  6  = 
cotangent  6,  abbreviated  cot  6  = 


1  =r 
sin  8  y 

1  =r 
cos  0~x' 

1        x 


tan0    y 


These  six  ratios  are  called  trigonometric  functions.  They 
are  of  the  greatest  importance  in  trigonometry  and  must 
be  learned  so  that  they  can  be  given  at  any  time  without 
hesitation. 

fA 
/ 

B 


Fio.  230. 


325.  Acute  angle  in  a  right  triangle. — In  the  right  triangle 

ABC,  Fig.  230,  sin  A=-  =  cos  B,  cos  A=-  =  sin  B,  tan  A  = 

c  c 

Of  O  C  C 

,  =  cot  B,  cot  A  =»  -  =  tan  B,  sec  A  «=  T  =  esc  B.  esc  A  =  -  =-  sec  B. 
b  a  b  a 

These  ratios  for  angle  B  will  readily  be  seen  if  the  triangle 
is  placed  in  the  position  of  Fig.  230(6). 

In  working  with  right  triangles,  the  following  forms  of  the 


TRIGONOMETRIC  FUNCTIONS 


407 


definitions  of  the  trigonometric  functions  will  be  found  con- 
venient for  either  acute  angle: 

side  opposite 


sin  A  (or  sin  J5)  = 
cos  A  (or  cos  Z?)  = 


hypotenuse 

side  adjacent 

hypotenuse 


D,      side  opposite 

tan  A  (or  tan  B)  —  -f-=-     ,.       —  > 

side  adjacent 

hypotenuse 

esc  A  (or  esc  B)  =  -~-         — r—  > 
side  opposite 

^      hypotenuse 
sec  A  (or  sec  B)  —  -^4 


side  adjacent 

,  _,.      side  adjacent 

cot  A  (or  cot  B)  =  -rj-          — r— • 

side  opposite 

A      B 


As  an  exercise  give  rapidly  the  functions  of  the  acute  angles 
of  the  right  triangles  shown  in  Fig.  231. 

326.  Relation  between  the 
functions  of  an  angle  and  the 
functions  of  its  complement. 
In  the  right  triangle,  angle 
A  +  angle  5  =  90°.  That  is, 
angle  A  and  angle  B  are  com- 
plements of  each  other.  (See 
Art.  90.)  In  the  previous 
article  it  is  seen  that  sin  A  = 
cos  B ;  cos  A  =  sin  B ;  tan  A  = 
cot  B;  etc.  From  this,  we  see  FlG<  232- 

that  any  function  of  an  angle  is  equal  to  the  co-function  of  the 
complement  of  the  angle.  For  example,  cos  30°==  sin  60°,  sec 
22°  =  esc  68°. 


10S 


PRACTICAL  MATHEMATICS 


327.  Trigonometric  functions  by  construction  and  measure- 
ment. —  Let  it  be  required  to  find  the  sine,  cosine,  and  tangent 
of  40°. 

Draw  an  angle  XOA=40°,  Fig.  232.  Take  a  convenient 
distance  OP  on  the  terminal  side  and  draw  the  perpendicular 
QP.  Then  by  measuring  the  lines  we  have  the  following: 


tan 


---  84 
~OQ~.96~ 


The  measurements  are  made  in  inches. 
As  an  exercise  draw  the  angles  and  fill  out  the  following 
table.     Carry  the  results  to  two  places  of  decimals. 


Angle 


tan 


Angle 


tan 


10° 


50° 


15° 


55° 


20° 


60° 


25° 


65° 


30° 


70° 


35° 


75° 


40° 


80° 


45° 


85° 


^ 


In  the  above  table,  compare  the  sine  and  cosine  of  10°  and 
80°,  15°  and  75°,  20°  and  70°,  etc.  As  in  Art.  326,  it  will  be 
noticed  that  any  function  of  any  angle  is  the  co-function  of 
the  complement  of  that  angle.  It  follows  that  any  function 
of  an  angle  larger  than  45°  is  a  function  of  some  angle  that  is 
less  than  45°.  If  then  a  table  is  made  for  the  functions  of  all 


TRIGONOMETRIC  FUNCTIONS 


409 


the  angles  from  0°  to  45°,  it  can  be  used  for  finding  the  func- 
tions of  the  angles  from  45°  to  90°  as  well.  Table  XI  is 
arranged  in  exactly  this  way.  It  includes  the  angles  for 
every  10'  from  0°  to  90°. 

328.  Use  of  functions  in  constructing  angles. — 

Example  1.     Construct  an  angle  of  40°. 

Construction,  sin  40°  =  0.64.  By  the  help  of  this,  the 
angle  may  be  drawn  as  follows:  Draw  a  straight  line  AB,  Fig. 
233.  At  some  point  Q  erect  a  perpendicular  QP  0.64  in.  in 
length.  With  P  as  a  center  draw  an  arc  with  a  radius  of  1  in. 
cutting  AB  at  0.  Draw  OP.  Angle  QOP  is  40°,  for  sin  QOP  = 
0.64  =  sin  40°. 


In  constructing  an  angle  of  40°  all  that  is  necessary  is  to 
make  the  sides  QP  and  OP  of  such  lengths  that  the  ratio  shall 
be  0.64.  For  this  1.28  in.  and  2  in.  could  be  used  conveniently. 
Usually  it  is  most  convenient  to  make  one  of  the  sides  unity. 
In  making  the  construction,  some  other  function  of  40°  could 
be  used  as  well  as  the  sine. 

Example  2.     Construct  an  angle  of  35°  by  using  tan  35°. 

Construction,  tan  35°  =  0.70.  Draw  a  straight  line  MN, 
Fig.  234.  At  some  point  Q  erect  a  perpendicular  QP  to  MN 
0.70  in.  in  length.  Locate  0,  making  OQ=l  in.  Draw  OP. 
Angle  QOP  is  35°,  for  tan  QOP  =  0.70  =  tan  35°. 

EXERCISES  112 

1.  Construct  by  the  use  of  sines,  angles  of  25°,  65°,  47°,  53°  20',  and 
25°  30'. 

Suggestion.     Use  data  from  the  table  of  Art.  327  or  from  Table  XI. 

2.  Construct  by  use  of  cosines,  angles  of  20°,  40°,  17°  20',  and  67°  40'. 

3.  Construct  an  angle  6  whose  tangent  is  2.     Find  the  other  functions 
of  this  angle. 

Ans.  sin  0  =  .89;  cos  0  =  .45;  cot  0  =  .50;  sec  6  =2.24;  esc  0  =  1.12. 


410 


PRACTICAL  MATHEMATICS 


4.  Construct  an  angle  0  whose  cosine  is  .8.     Find  the  other  functions 
of  this  angle. 

Ans.  sin  0-.6;  tan  0-.75;  cot  0-1.33;  sec  0-1.25;  esc  0-1.67. 

5.  Construct  0  when  (a)  sec  0-3;    (b)  cot  0-5;    (c)  esc  0-2;  (d) 
sin  0  =  .3. 

329.  Values  of  functions  by  computa- 
tion.— From  our  knowledge  of  geometry 
we  know  the  relations  between  the  sides 
and  the  angles  of  right  triangles  when  the 
1 "  acute  angles  are  30°,  45°,  or  60°. 

Thus,  when  the  acute  angles  are  45°  each,  the 
two  legs  are  equal;  and  if  they  are  given  the 
hypotenuse  can  be  found.  If  one  acute  angle 
is  30°  and  the  other  60°,  the  shorter  leg  is  one 
half  the  hypotenuse. 

(1)  The   45°    angle.     Draw    a    right    triangle   ABC,    Fig. 
235,  with  the  angle  4=45°.     Then  the  angle  5  =  45°.     Let 
'  =  land  (75  =  1.     Then  45  =  VlHT2  =  A/2. 

'     '-,=  0.707, 


sin  45°  = 


cos  45°  = 


V2 
1 


V2 
tan  45°  =  1. 

(2)  The  60°  and  30°  angles.  Draw  a 
right  triangle  with  the  acute  angles  60° 
and  30°  respectively,  Fig.  236.  Let 
AC  =  1,  then  AB  =  2  and  C5  =  \/3- 
The  functions  can  now  readily  be 
written: 

sin  60°  =  ^?  = 


sin  30°  = 


cos  60°  = 


tan  60°  = 


tan 


--- 

V3 


330.  Angles  in  other  quadrants. — In  the  second  quadrant, 
the  angle  XOA  =  120°,  Fig.  237,  has  the  same  numerical  ratios 
between  the  abscissa,  ordinate,  and  distance  of  the  point  P 


TRIGONOMETRIC  FUNCTIONS 


411 


in  the  terminal  side  as  the  ratios  between  the  abscissa,  ordinate, 
and  distance  of  the  point  P'  in  the  terminal  side  of  the  angle 

'  =  60°  in  the  first  quadrant. 
Thus,  using  the  values  given  in  the  figure, 

sin  120°  =  ~~  and  sin  60°       =~, 

cos  120°  =  -  i  and  cos  60°      =  £, 
tan  120°  =  -  \/3  and  tan  60°  =  Vs. 

y 

B 


FIG.  237. 

There  are  angles  in  the  second,  third,  and  fourth  quadrants 
whose  functions  are  connected  in  this  way  with  the  angles  30°, 
45°,  and  60°  in  the  first  quadrant.  What  are  they? 

The  functions  of  these  angles  can  easily  be  written  by 
remembering  (1)  that  the  distance  is  always  positive;  (2)  that 
the  ordinate  is  positive  in  the  first  and  second  quadrants  and 
negative  in  the  third  and  fourth;  (3)  that  the  abscissa  is  posi- 
tive in  the  first  and  fourth  quadrants  and  negative  in  the  second 
and  third.  (See  Art.  286.) 

331.  Angles  of  90°,  180°,  270°,  and  0°.— For  an  angle  of  90°, 
the  ordinate  equals  the  distance  and  the  abscissa  is  zero. 
The  functions  are  as  follows : 


sin  90°  =^  =  1, 

esc  90°  =  -  =  1, 

r 

y 

cos  90°  =  -  =  0, 

sec  90°  =  -=  co 

r 

X 

tan  90°  --=0°, 

cot  90°  =  -  =  0. 

X 

y 

412 


PR  A  CTICAL  MA  TH  EM  A  TICS 


The  symbol  »  is  read  infinity,  and  here  means  that  as  the 
angle  increases  to  90°,  the  tangent  and  the  secant  increase 
without  limit. 

That  the  values  of  the  functions  are  as  given  will  readily 
be  seen  if  one  draws  an  angle  XOP,  Fig.  238,  nearly  equal 
to  90°,  and  considers  the  values  of  the  ratios  as  the  angle 
changes  to  90°. 

For  an  angle  of  180°,  the  ordinate  is  zero  and  the  abscissa 
and  distance  are  equal.  The  functions  are  as  follows: 


sin  180°  =  -=0, 

r 

cos  180°  =  *=  -1, 
tan  180°  =  ^  =  0, 


esc  180°=    =co 

y 

sec  180°  =  -= -1. 

x 

cot  180°  =  -=«. 

y 


These  values  may  be  found  by  considering  angle  XOR, 
Fig.  238. 

Y 


Fio.  238. 

Similarly  the  functions  for  270°  and  0°  are  as  follows: 

esc  270° 
sec  270° 
cot  270C 
cscO° 
secO° 
cotO° 


=  0, 

=   00, 

—  1 
A> 


sin  270°  =-1, 
cos  270°  =0, 
tan  270°  =  oo , 
sin  0°      =0, 
cos  0°      =  1, 
tanO°     =0, 

332.  The  sine,  cosine,  and  tangent  of  each  of  the  angles 
mentioned  in  the  previous  articles  are  arranged  in  the  follow- 
ing table.  The  student  should  carefully  verify  each  result  by 
drawing  a  figure  and  computing  the  ratios.  Also  express  the 


TRIGONOMETRIC  FUNCTIONS 


413 


results  in  decimals  and  compare  with  the  values  given  in 
Table  XI. 


Deg. 

Rad. 

sin 

cos 

tan 

Deg. 

Rad. 

sin 

cos 

tan 

77T 

V3 

1 

0° 

0 

0 

1 

0 

210° 

IT 

-\ 

V3 

30° 

7T 

6 

i 

V'3 
2 

1 

V3 

225° 

57T 

4 

1 

~V2 

1 

~V2 

1 

45° 

IT 

4 

1 

V2 

1 

V2 

1 

240° 

47T 

3 

V3 
2 

~  2 

V/3 

60° 

7T 

3 

IT 

1 

V3 

270° 

2 

-1 

0 

00 

90° 

X 

2 

1 

0 

oo 

300° 

STT 

3 

V3 
2 

-2 

-V3 

120° 

27T 

3 

V3 
2 

-1 

-Vs 

315° 

77T 

4 

1 

1 

V2 

-1 

135° 

STT 
4 

1 

V2 

1 

V2 

-1 

330° 

UTT 

6 

i 

2 

V3 
2 

1 

150° 

STT 
6 

? 

V3 
2 

1 

360° 

27T 

0 

1 

0 

180° 

IT 

0 

-1 

0 

CHAPTER  XXXVII 
TABLES  AND  THEIR  USES 

333.  Nature  of  trigonometric  functions. — From  what  has 
been  done  with  the  trigonometric  ratios,  it  is  seen  that  they  are 
abstract  numbers  and,  in  general,  cannot  be  expressed  exactly 
as  decimals.     For  convenience  in  computing,  these  ratios  are 
arranged  in  tables  somewhat  similar  to  the  tables  of  logarithms. 
The  ratios  in  these  tables  may  be  carried  to  any  number  of  deci- 
mal places.     The  larger  the  number  of  decimal  places,  the  more 
nearly  accurate  the  computations  with  the  tables  will  be. 

334.  Table  of  functions. — In  Table  XI  are  arranged   the 
natural  and  logarithmic  functions  of  angles  for  every   KX 
from  0°  to  90°.     The  logarithms  have  10  added  when  the 
characteristics  are  negative  to  avoid  the  writing  of  negative 
signs  in  the  table.     These  logarithms  are  simply  the  logarithms 
of  the  natural  functions  that  are  in  the  adjoining  columns,  and 
are  placed  here  for  convenience. 

Since,  as  stated  in  Art.  327,  each  acute  angle  above  45°  has 
as  a  function  the  co-function  of  an  angle  less  than  45°,  each 
number  in  the  table  serves  as  the  function  of  two  different 
angles  whose  sum  is  90°. 

The  angles  less  than  45°  are  found  at  the  left  of  the  page, 
and  the  names  of  the  functions  at  the  top  of  the  page.  The 
angles  greater  than  45°  are  found  at  the  right  of  the  page,  and 
the  names  of  the  functions  at  the  bottom  of  the  page. 

335.  To  find   the  function  of  an  angle  from  the  table.— 
To  find  the  function  of  an  angle  from  the  table  we  proceed 
much  the  same  as  with  the  table  of  logarithms.     It  can  best 
be  illustrated  by  examples. 

(A)  When  the  angle  is  given  in  the  table. 
Example  1.     Find  the  tangent  of  23°  20'. 
Find  the  angle  of  23°  20'  at  the  left  of  the  page  and  read 
0.4314  in  the  column  headed  natural  tangent, 
/.  tan  23°  20' =  0.4314. 
414 


TABLES  AND  THEIR  USES  415 

Example  2.     Find  the  cosine  of  86°  40'. 
Find  86°  40'  at  the  right  of  the  page  and  read  0.0581  in  the 
column  with  natural  cosine  at  the  bottom, 

.'.  cos  86°  40' =  0.0581. 

(B)     When  the  angle  is  not  given  in  the  table. 

Example  3.     Find  sin  17°  27'. 

Find  sin  17°  20' =  0.2979.  Find  the  tabular  difference 
between  0.2979  and  the  next  ratio  below.  This  difference  is 
0.0028.  Since  this  difference  is  for  10',  the  difference  for  7' 
is  0.7X0.0028  =0.0020, 

.*.  sin  17°  27' =  0.2979+0.0020  =  0.2999. 

Note  that  the  interpolation  is  very  similar  to  that  in  loga- 
rithms, and  gives  approximate  results  only. 
Example  4.     Find  tan  69°  43.6'. 
Find  tan  69°  40' =  2.6985. 
Tabular  difference  for  10' =  0.0243. 
Difference  for  3.6' =  0.0243X0.36  =  0.0087. 

.'.  tan  69°  43.6' =  2.7072. 

Example  5.     Find  cos  37°  57.3'. 
Find  cos  37°  50' =  0.7898. 
Tabular  difference  for  10' =  0.0018. 
Difference  for  7.3' =  0.0018X0.73  =  0.0013. 

.*.  cos  37°  57.3' =  0.7898 -0.0013  =  0.7885. 

It  is  to  be  noted  that  a  subtraction  is  to  be  made  when  inter- 
polating in  finding  a  cosine  or  a  cotangent  of  an  angle;  while 
in  finding  a  sine  or  a  tangent,  an  addition  is  performed.  This 
is  because  as  the  angle  increases  from  0°  to  90°  the  sine  and  the 
tangent  increase,  but  the  cosine  and  the  cotangent  decrease. 

336.  To    find  the  angle    corresponding  to  a  function. — 

(A)  When  the  function  is  given  in  the  table. 

Example  1.     Find  x  if  sin  x  =  0.2728. 

Find  0.2728  in  the  column  labeled  natural  sine  and  read 
15°  50'  in  the  column  labeled  angle. 

.'.  x  =  15°  50'. 


416 


PRACTICAL  MATHEMATICS 


(B)  When  the  function  is  not  given  in  the  table. 

Example  2.     Find  x  if  tan  x  =  1.5725. 

In  the  column  labeled  natural  tangent,  find  the  ratio  nearest 
1.5725  and  smaller.  This  is  1.5697  and  is  the  tangent  of 
57°  30'.  The  tabular  difference  is  0.0101.  The  difference  be- 
tween 1.5697  and  1.5725  is  0.0028.  Since  a  difference  of  10' 
gives  a  difference  in  the  ratio  of  0.0101,  it  will  take  a  difference 
of  as  many  minutes  to  give  a  difference  in  the  ratio  of  0.002$ 


/.  x  =  57°  30'+2.8'  =  57°  32.8'. 

Note  again  how  similar  the  interpolating  is  to  that  in 
logarithms. 

Example  3.     Find  x  if  cos  x  =  0.7396. 

Since  the  cosine  decreases  as  the  angle  increases,  find  the 
cosine  nearest  to  .7396  but  larger.  This  is  cos  42°  10'  =  0.7412. 

Tabular  difference  =0.0020. 

Difference  of  0.7412-0.7396  =  0.0016. 


'  =  8'. 

z  =  42010'+8'=42°18'. 
Example  4.     Find  x  if  log  sin  x  =  9.3762  -  10. 
From  table  log  sin  13°  40'  =  9.3734. 
9.3762-9.3734  =  0.0028. 
Tabular  difference  =  0.0052. 

MX  10'  =  5.4'. 
;.  z  =  13°40'+5.4'  =  13°  45.4'. 


EXERCISES  113 

1.  Find  the  sine,  cosine,  and  tangent  of  40°  10',  59°  50',  76°  30', 
and  5°  40'. 

2.  Find  the  sine,  cosine,  and  tangent  of  (a)  17°  36',  (b)  29°  29',  (c) 
76°  14',  (d)  83°  33',  (e)  63°  47'. 


Ana.  (a) 


0.3024 
0.9532  (b) 
0.3172 


0.4922 
0.8705  (c) 
0.5654 


0.9713  [  0.9937 

0.2380  (d)   |  0.1123  (e) 
4.0817  8.8468 


0.8971 
2.4418 
2.0308 


3.  Find    the    angles    having  the  following  as  sines:  0.5807,   0.2725, 
0.4986,  0.9127,  0.0276. 

Ans.  35°  30',  15°  48.9',  29°  54.4',  65°  52.7',  1°  34.8'. 

4.  Find  the  angles  having  cosines  as  follows:  0.3764,  0.8642,  0.9091, 
0.4848,  0.0986.  Ans.  67°  53.3',  30°  12.7',  24°  37.5',  61°,  84°  20.5. 


TABLES  AND  THEIR  USES  417 

6.  Find  the  angles  having  tangents  as  follows:  0.2256,  1.7624,  2.8427, 
0.1111,  3,  0.6666. 

Ans.  12°  42.9',  60°  25.7',  70°  37.1',  6°  20.3',  71°  33.9',  33°  41.2'. 

6.  Find  sin  34°  40'  and  find  the  logarithm  of  this  result  from  Table  X. 
Find  log  sin  34°  40'  from  Table  XI  and  compare  results. 

7.  Verify  the  following  by  the  tables: 

log  sin  56°  35' =  9.9215;  log  tan  34°  15.6' =  9.8332; 
log  cos  27°  55'  =9.9462;  log  cos  19°  53.4' =9. 9733; 
log  sin  17°  9'  =  9.4696;  log  tan  75°  56.8'  =0.6015. 

8.  Find  x  in  each  of  the  following: 

(a)  log  cos  z  =  9.8236;  (c)  log  tan  x  =  0.4293; 

(b)  log  sin  x  =  9.4737;  (d)  log  cot  x  =  9.4236. 

Ans.  (a)  48°  13.6';  (b)  17°  19';  (c)  69°  35.3';  (d)  75°  8.8'. 

337.  Evaluation  of  formulas. — Formulas  in  various  lines 
of  work  often  contain  trigonometric  functions.  As  with  other 
formulas,  these  can  usually  be  evaluated  with  or  without 
logarithms.  Since  logarithms  are  a  very  convenient  and 
useful  tool,  they  should  be  used  whenever  they  can  be  used 
to  advantage. 

To  indicate  the  power  of  a  trigonometric  function  the  ex- 
ponent is  placed  before  the  symbol  for  the  angle. 

Thus,  sin2  30°  means  the  square  of  sin  30°. 

The  logarithms  of  trigonometric  functions  are  found  in 
Table  XI  and  those  of  numbers  in  Table  X. 

Example  1.     Find  \Xsin  47°+tan3  36°. 

Solution.  sin  47°  =0.7314,  from  Table  XI. 

log  tan  36°  =  9.8613 -10. 
/.  log  tan3  36°  =  9.5839 -10. 

/.  tan3  36°  =  0.3836,  from  Table  X. 
.".  sin  47°  + tan3  36°  =  0.7314 +0.3836  =  1.1150. 

log  1.1150  =  0.0473. 
log  \/l.  1150  =  0.0158. 
.'.  \/l.  1150  =  1.037. 
.'.  \Xsin  47°+tan3  36°  =  1.037.  Ans. 

Example  2.     Given  x  =  —     0      ,    >  find  the  value  of  x  to 

O  t7       ^.\J 

two  decimal  places. 

In  an  example  like  this  it  is  agreed  that  69°  40'  shall  be 
changed  to  radians  to  give  the  number  to  divide  by. 

Solution,     tan  72°  34'  =  3. 1846. 

27 


418  PRACTICAL  MATHEMATICS 

69°  40'  =  69$°  =  69§X0.01745  radians  =  1.216  radians. 

3.1846^-1.216  =  2.62. 

/.  z  =  2.62.  Ana. 

Logarithms  could  be  used  in  solving  this  example. 

EXERCISES  114 


1.  Find  the  value  of  Vsin*  49°  10'.  Ans.  0.6583. 

2.  Find  the  value  of  \/tan  75°  +56.  Ans.  2.266. 

3.  Find  the  numerical  value  of  r$  (s2—  <2)  tan  0,  where  r  =  25.2,  a  =  90, 
*=49.6,  and  tf  =  31°  52'.  Ans.  30,140. 

4.  Find  the  value  of  ae-*  sin  (,d+0),if  a  =  5,  6=200,c  =  600,  0=  -0.1745 
radians,  e  =2.718,  and  t  =0.001.  Ans.  1.69. 

Suggestion,     (ct  +  0)  =0.6-}-  (-0.1745)  =0.4255  radians. 
0.4255X57.2957°  =24.379°  =24°  22.7'. 

sin  45°  56'  20"  _ 

6.  Given  x  —  -  „„„  on/  -  >  find  the  value  of  x  to  three  decimal  places. 
oO    £\J 

Suggestion.     First  change  36°  20'  to  radians.  Ans.  1.133. 

6.    Given  x  =  —  —  >  find  x  to  four  decimal  places. 


_  Ans.  0.8689. 

7.  Evaluate  \/o2+b2-2a6cos  C,  when  a  =  231,  6=357,  and  C  =  55°. 

^rts.293.6. 

8.  Evaluate  q  sm  B  *m  C,  when  a  =  126,  A  =30°,  B  =  72°,  C  =  78°. 

sin  A 

Ans.  234.4. 

9.  Find  the  value  of  sin  x  cos  y+  cos  x  sin  j/,  when  z  =  42°  10',  and 
y  =  17°  50'.  Ans.  0.866. 

10.  Find  the  value  of  each  of  the  following: 

(a)  sin2  20°+cos2  20°.  (c)  sin2  40°+cos2  40°. 

(b)  sin2  30°+cos2  30°.  (d)  sin2  62°  30'+cos2  62°  30'. 
Compare  the  results  in  the  above  and  state  conclusions. 

11.  The  velocity  t;  of  a  body 
sliding  a  distance  s  down  a  smooth 
plane  inclined  at  an  angle  <p  with 
the  horizontal  is  given  by  the 
formula  v  =  \/20s  sin  <p,  where 
0  =  32.  Find  v  when  s  =  50  ft. 
and  *>  =  27°  16'. 

Ans.  38.3  ft.  per  second. 

12.  If  the  resistance  of  the  air  is  disregarded,  the  distance  along  a 
horiz^  ntal  plane  that  a  projectile  will  go  is  given  by  the  formula  d  = 

-  •  ,  where  v  is  the  velocity  at  which  the  body  is  projected  in  feet 

if 

per  second,  a  the  angle  that  the  initial  direction  makee  with  the  hori- 
zontal, and  d  the  distance  along  the  horizontal.     The  value  of  g  may  be 


TABLES  AND  THEIR  USES 


419 


taken  as  32.  Find  d if  vis  800ft.  per  second  and  a  is  5°.  Using  the  same 
velocity,  find  d  when  a.  is  20°,  30°,  40°,  and  45°. 

Ans.  3472  ft.;  12,856  ft.;  17,320  ft.;  19,696  ft.;  20,000  ft, 
13.  Disregarding  the  resistance  of  the  air,  the  highest  point  reached  by 

4j2     gl  ri    2      fy 

a  projectile  is  given  by  the  formula  y= ~ Find  the  greatest 

^9 

height  above  the  starting  point  reached  by  a  projectile  having  an  initial 
velocity  of  2000  ft.  per  second,  and  having  successive  values  for  a  of  5°, 
10°,  20°,  30°,  45°,  60°,  and  90°. 


FIG.  240. 

14.  The  height  y  that  a  projectile  is  after  traversing  a  horizontal  dis- 
tance x,  when  projected  with  a  velocity  v  in  a  direction  making  an  angle 
a  with  the  horizontal,  is  given  by  the  following  formula : 

gx2 


y—xt&a  a—- 

2v2  COS-  a 

Find  y  when  a;  =  1000  yd.,  y  =  2000  ft.  per  second,  a  =  5°,  and  0  =  32. 

Ans.  226.2ft. 

15.  If  the  resistance  of  the  air  is  disregarded,  the  greatest  horizontal 
distance  a  projectile  will  go  is  found  by  making  the  initial  direction  at 
an    angle    of    45°    with    the    horizontal. 

Find  the  greatest  horizontal  distance 
that  a  shell  having  an  initial  velocity  of 
2200  ft.  per  second  can  reach. 

Ans.   28.6+  miles. 

16.  If  F  is  the  force  required  to  move 
a  weight  W  up  a  plane  inclined  to  the 
horizontal  at  an  angle  a,  and  M  (Greek 
letter  mu)  the  coefficient  of  friction,  then 


Source 


sin  a+M  cos  a 
cos  a—  n  sin  d 


FIG.  241. 


Calculate  F  if  TF  =  800  lb.,  a  =  30°,  and  M  =  0.2.  Ans.  703  Ib. 

17.  In  computing  the  illumination  on  a  surface  when  the  surface  is 
not  perpendicular  to  the  rays  of  light  from  a  source  of  light,  the  following 
formula  is  used: 

/ 


420 


PRACTICAL  MATHEMATICS 


where       /-.'the  illumination  at  the  point  on  the  surface  in  foot-candles 
/  =  the  luminous  intensity  of  the  source  in  candles, 
rf  =  the  distance  in  feet  from  the  source  of  light, 
<?  =  the  angle  between  the  incident  ray  and  a  line  perpendicular 
to  the  surface. 

Solve  this  formula  for  d  and  /,  and  obtain  the  following  formulas: 


//  cos 


Ed* 


18.  By  means  of  the  preceding  formulas  compute: 

(1)  E  when  7=50,  d  =     10,  and  *>  =  75°. 

(2)  d  when  7  =  60,  E  =  0.25,  and  <p  =  65°. 

(3)  7when#=  4,  d=       8,  and  *>  =  45°. 

Ans.  (1)  0.1294;  (2)   10ft.;  (3)  362. 

19.  What  do  the  formulas  in  exercise  17  become  if  ^>=0°?     That  is, 
if  the  rays  are  normal  (perpendicular)  to  the  surface. 

Source  5 


Horizontal 


Fio.  242. 

20.  To  compute  the  illumination  on  a  horizontal  surface  from  a  source 
of  light  at  a  given  vertical  distance  from  the  surface  the  following  formula 
is  used: 


where  Eh=the  illumination  in  foot-candles  at  a  point  on  the  horizontal 

surface, 

/  =the  luminous  intensity  of  the  source  in  candles, 
h=ihe  vertical  distance  in  feet  from  the  horizontal  surface  to 

the  source  of  light, 

v>  =  the  angle  between  the  incident  ray  and  a  vertical  line. 
Solve  this  formula  for  h  and  /,  and  obtain  the  following  formulas: 


21.  By  means  of  the  preceding  formulas  compute: 
(l)^when    7=250,     h  =  12,  and  ^  =  55°. 

(2)  h  when     /  =  100,   Eh  =  65,  and  ?  =  12°. 

(3)  /  when  Eh  =0.85,    h  -  8,  and  *  =  37°. 

Ans.  (1)  0.3276;  (2)  1.2  ft.;  (3)   106.8. 


CHAPTER  XXXVIII 
RIGHT  TRIANGLES 

338.  Any  triangle  has  three  sides  and  three  angles;  these  are 
called  the  six  elements  of  the  triangle. 

The  angles  are  usually  represented  by  the  capital  letters  A, 
B,  and  C;  the  sides  by  the  small  letters  a,  b,  and  c,  the  side 
a  being  opposite  angle  A,  side  b  opposite  angle  B,  and  side  c 
opposite  angle  C. 

To  solve  a  triangle  is  to  find  the  values  of  the  remaining 
elements  when  some  of  them  are  given. 

339.  Solving. — A  triangle  may  be  solved  in  two  ways : 

(1)  By  constructing  the  triangle  from  the  known  elements, 
and  measuring  the  remaining  elements  with  the  ruler  and  the 
protractor. 

(2)  By  computing  the  remaining  elements  from  those  that 
are  known. 

The  first  has  already  been  done  to  some  extent  in  Chapter 
XIII.  The  second  has  been  done  for  some  special  triangles, 
as  the  right  triangle,  the  isosceles  triangle,  and  the  equilateral 
triangle,  in  Chapter  XI,  but  only  for  some  of  the  elements,  not 
including  the  angles. 

By  trigonometry  a  triangle  can  always  be  solved  when  the 
facts  given  are  sufficient  for  its  construction;  and  not  only  can 
the  sides  be  found,  but  the  angles  also. 

EXERCISES  115 

If  A,  B,  and  C  represent  the  angles  of  a  triangle,  and  a,  b,  and  c  re- 
spectively the  sides  opposite  these  angles,  construct  carefully  the  follow- 
ing triangles,  to  scale  if  necessary,  and  measure  the  other  elements.  It 
is  important  that  the  student  should  carry  out  these  carefully  for  it  will 
help  him  to  see  when  there  are  sufficient  data  for  solution. 

1.  A  =40°,  b  =2  in.,  c  =  2.5  in.,  find  B,  C,  and  a. 

2.  .4=50°,  C  =  70°,  b  =  2  in.,  find  B,  a,  and  c. 

3.  A  =30°,  a  =  10  ft.,  c  =  15  ft.,  find  B,  C,  and  6. 

421 


422  PRACTICAL  MATHEMATICS 

Can  more  than  one  triangle  be  formed  from  the  data  in  3? 

4.  a  =  20  ft.,  6  - 15  ft,  c  =  12  ft.,  find  A,  B,  and  C. 

5.  A  =40°,  £=80°,  C-60°,  find  a,  b,  and  c. 

Can  more  than  one  triangle  be  formed  from  the  data  in  5? 
The  following  are  right  triangles  and  C  is  the  right  angle: 

6.  A  =29°,  6=2  in.,  find  B,  a,  and  c. 

7.  A  =42°,  a  =  4  in.,  find  B,  b,  and  c. 

8.  A  =47°,  c  =  3  in.,  find  B,  a,  and  c. 

9.  a  =4  in.,  &  =  6  in.,  find  A,  B,  and  c. 
10.  a  -1.5  ft.,  c  =  2.3  ft.,  find  A,  B,  and  6. 

340.  The  right  triangle. — The  right  triangle  has  already 
been  solved  when  any  two  sides  are  known,  but  the  angles 
were  not  found.     The  previous  exercises  from  6  to  10  should 
lead  us  to  expect  that  we  could  find  the  other  elements  when 
any  two  are  given,  other  than  the  right  angle,  and  including  at 
least  one  side.     Geometry  will  not  do  this  but  trigonometry  will. 

It  is  well  to  recall  the  following  facts  concerning  the  right 
triangle : 

(1)  The  hypotenuse  is  greater  than  either  of  the  other  two  sides, 
and  less  than  their  sum. 

(2)  The  square  of  the  hypotenuse  is  equal  to  the  sum  of  the 
squares  of  the  other  two  sides. 

(3)  The  sum  of  the  two  acute  angles  is  90°,  that  is,  the  acute 
angles  are  complements  of  each  other. 

(4)  The  greater  side  is  opposite  the  greater  angle,  and  the 
greater  angle  is  opposite  the  greater  side. 

An  inspection  of  the  problems  of  construction  will  show  that 
all  the  possible  sets  of  two  given  parts  for  the  right  triangle 
are  included  in  the  following  cases: 

CASE  I.    Given  an  acute  angle  and  a  side  not  the  hypotenuse. 

CASE  II.    Given  an  acute  angle  and  the  hypotenuse. 

CASE  III.    Given  the  hypotenuse  and  one  other  side. 

CASE  IV.    Given  the  two  sides  not  the  hypotenuse. 

341.  Directions  for  solving. — To  solve  a  right  triangle  it  is 
necessary  that  two  elements  be  given,  at  least  one  of  which  is  a 
side. 

Each    equation,   as    sin.A=->    contains   three   quantities. 

C 

When  two  of  these  are  given  the  third  can  be  found.     These 
equations  together  with  the  facts  from  geometry:   (1)  that 


RIGHT  TRIANGLES 


423 


the  square  of  the  hypotenuse  equals  the  sum  of  the  squares  of 
the  other  two  sides;  and  (2)  that  the  sum  of  the  two  acute 
angles  equals  90°,  enable  one  to  solve  any  right  triangle. 
These  equations  may  be  written  thus  : 

(1)  sin  A=a> 

(2)  cos  A=-, 

c 

(3)  tan  A  =j-> 

(4)  cot  A=-, 

a 

(5)  c2  = 

(6)  sin 


FIG.  243. 


(7)  cos 

(8)  tan 


(9)  COt  B  =  ^, 

(10)  .4 +5  =  90°. 


Notice  that  all  of  these  except  (5)  and  (10)  are  nothing  but 
the  definitions  of  the  trigonometric  ratios. 

342.  Case  I.    Given  A  and  b,  A  and  a,  B  and  a,  or  B  and  b.— 

Example.  In  a  right  triangle  ,4=32°  20'  and  &  =  10  ft., 
find  B,  a,  and  c. 

Solution.  First,  construct  the  triangle  carefully;  second, 
write  equations  using  two  of  the  known  elements  and  one  of 
the  unknown  in  each. 


Equations. 


Construction. 


(1)  A  +£  =  90°,  this  gives  B. 

(2)  v-o.s  A  =->  this  gives  c. 

C 

(3)  tan  A  —  j->  this  gives  a. 


Substituting  in  (1),  32°  2 

.-.  B  =  90°  -32°  20'  =57°  40'. 


424  PRACTICAL  MATHEMATICS 

Substituting  in  (2),  cos  32°  20'  =—- 

c 

.'.  c  =  10-!- cos  32°  20'  =  10 -K8450  =  11. 83  ft. 
Substituting  in  (3),  tan  32°  20'  =  y^ 

.'.  a=  lOXtan  32°  20'  =  10 X .6330  =  6.33  ft. 

This  may  be  checked  (a)  by  measuring  the  elements  in  the 
triangle  constructed;  (b)  by  using  some  other  equation  than 
the  ones  used  in  solving. 

Thus,  substitute  values  in  c2  =  o2+62. 
11.832  =  6.332+102. 
139.95  =  39.97  +  100  =  139.97,  which  agrees  closely. 

This  solution  has  been  carried  through  with  natural  func- 
tions. Logarithms  could  be  used  to  advantage  in  performing 
the  multiplications  and  divisions. 

Thus,  in  (2)  c  = -A 

cos  A 

.'.  log  c  =log  b— log  cos  A. 

log  10  =  1.0000 
log  cos  32°  20'  =  9.9268  - 10 


log  c=  1.0732 

.'.  c=11.83+ft.,  which  agrees  with  the  result  ob- 
tained before. 

343.  Directions  for  solution  of  triangles. — 

(1)  Construct  the  triangle  as  accurately  as  possible  with  in- 
struments.    This  gives  a  clear  idea  of  the  relation  of  the  parts, 
and  will  detect  any  serious  blunder  in  computation. 

(2)  Write  dawn  all  the  equations  necessary  to  find  the  elements 
wanted. 

(3)  Compute  the  elements  by  natural  or  logarithmic  functions. 

(4)  Check  the  work. 

(5)  Strive  for  neatness  and  clearness  in  the  work. 


RIGHT  TRIANGLES 


425 


344.  Case  II.     Given  A  and  c  or  B  and  c.— 

Example.     Given  A  =  67°  42.8'  and  c  =  23.47  ft. 

Formulas.  Construction. 

Q0°.       '    B  =  Q(r  —  A.  B 


(2)  win  A=    ,         .' .   d  =  csmA. 


(3)cosA=-j        .'.    b  =  ccosA. 
c 

Computation  by  logarithms. 


6742.8 


By  (1),  B  =  90° -67°  42.8'  =  22°  17.2'. 
By  (2),    a  =  23.47  sin  67°  42.8'. 
By  (3),  6  =23.47  cos  67°  42.8'. 

log  23.47  =  1.3705  log  23.47  =  1.3705 

log  sin  67°  42.8'  =  9.9663      log  cos  67°  42.8'  =  9.5789 


b 

FIG.  245. 


log  a  =  1.3368 
/.  o  =  21.72 


log  6  =  0.9494 
.'.  6  =  8.90 
Check.     Usinga2  =  c2-62=(c+6)(c-6). 
log  (c+6)  =  1.5101 
log  (c-6)  =  1.1635 

log  (c2-62)  =2.6736  =  log  a2. 
345.  Case  III.     Given  c  and  a  or  c  and  b.— 
Example.     Given  c  =  35.62  ft.  and  a  =  23.85  ft.,  find  6,  A,  B. 

Formulas.  Construction. 

(1)  sin  A=- 


'.  A  =sin~l 

(2)  cos  B  = 

(3)  tan  A  = 


426  PRACTICAL  MATHEMATICS 

Computation. 

log  23.85  =  1.3775  log  23.85  =  1.3775 

log  35.62  =  1.5517  log  tan  A  =  9.9549 

log  sin  A  =  9.8258  =  log  cos  B  log  b  =  1 .4226 

.'.  A  =42°  2.1' and  5=  47°  57.9.'  .'.  6  =  26.46. 

Check. 

log  (c+6)     =1.7930 
log  (c-6)     =0.9619 


log  (c2-62)  =2.7549  =  log  a2. 

346.  Remark  on  inverse  functions.  —  The  form  yl=sin~l  - 

c 

a  " 
is  read  "A  =  the  angle  whose  sine  is  -•    This  is  a  convenient 

C 

way  of  expressing  the  fact,  and  allows  the  angle  symbol  to 
stand  alone.  The  —  1  that  is  in  the  position  of  an  exponent 
is  not  a  negative  exponent  in  meaning. 

The  form  sin"1  -  is  called  an  inverse  trigonometric  function. 
c 

It  is  also  written  arcsin  -  and  invsin  —     These  forms  are  also 

c  c 

.  .  .      a  »  ,<        .      a  »       ,  ,,.  .      a  » 

read     antisine  -»       arcsine  ->    and     inverse  sine  -• 
c  c  c 

347.  Case  IV.     Given  a  and  b.— 
The  equations  are: 


(1)  c2  =  o2 

(2)  tan  A  =r>  .'.A  =  tan-1  T- 

=  -,  /.  B  =  tan~1-- 

a  a 

Check,     sin  A  =  -•>  and  cos  A  =-• 
c  c 

EXERCISES  116 

Solve  the  following  right  triangles  and  check  each  by  making  an 
accurate  construction  and  by  substituting  into  a  formula  not  used  in 
solving. 

1.  A  =27°  30',  a  =  14  in.  ;  find  B,  b,  and  c. 

An*.  £-62°  30';  6-7.288;  r  =  15.78. 


RIGHT  TRIANGLES  427 

2.  B=46°  25',  a  =  17  ft.;  find  A,  b,  and  c. 

Ans.  A  =43°  35';  6  =  17.86;  c  =24.66. 

3.  A  =75°  26',  6  =  25  ft.;  find  B,  a,  and  c. 

Ans.  5  =  14°  34';  a  =  96.2;  c  =  99.38. 

4.  5  =  62°  40',  6=2  ft.;  find  A,  a,  and  c. 

5.  A  =  17°  50',  c=47  yd.;  find  B,  a,  and  b. 

6.  5=53°  20',  c  =  21  ft.;  find  A,  a,  and  6. 

7.  a  =  2  ft.,  c  =  3  ft.;  find  A,  B,  and  b. 

8.  6  =4  ft.,  c  =  9  ft.;  find  A,  B,  and  a. 

9.  a  =  4.23  in.,  6  =  7.23  in.;  find  A,  B,  and  c. 

10.  a  =27  in.,  6  =20  in. ;  find  A,  B,  and  c. 

11.  5=29°  45',  c  =  2.36  ft.;  find  A,  a,  and  6. 

12.  A  =32°  12',  c  =  8.23  in.;  find  B,  a,  and  6. 

348.  Orthogonal  projection. — If  from  a  point  P,  Fig.  247 (a), 
a  perpendicular  PQ  be  drawn  to  any  straight  line  RS, 
then  the  foot  of  the  perpendicular  Q  is  said  to  be  the  orthogonal 
projection  or  simply  the  projection  of  P  upon  RS. 


Olx^  I       \^ 

\^£§_ !    A 

A  s     \  E1 

FIG.  247. 

The  projection  of  a  line  segment  upon  a  given  straight  line 
is  the  portion  of  the  given  line  lying  between  the  projections 
of  the  ends  of  the  segment. 

In  Fig.  247(6)  and  (c),  CD  is  the  projection  of  AB  upon 
OX.  In  each  case  AE  —  CD  and  AE  =  AB  cos  9.  Hence,  if 
I  is  the  length  of  the  segment  of  the  line  projected,  p  the  pro- 
jection, and  Q  the  angle  between  the  lines,  then 

p  =  l  cos  0. 

Similarly,  the  projection  of  AB  upon  a  line  OY  that  is  per- 
pendicular to  OX  and  in  the  same  plane  as  OX  and  AB  is 

p'  =  l  sin  6. 

349.  Vectors. — In  physics  and  engineering,  line  segments 
are  often  used  to  represent  quantities  that  have  direction  as 
well  as  magnitude.  Velocities,  accelerations,  and  forces  are 
such  quantities. 


428 


PRACTICAL  MATHEMATICS 


R 


0 


Q 


FIG.  248. 


For  instance,  a  force  of  100  Ib.  acting  in  a  northeasterly 
direction  may  be  represented  by  a  line,  say,  10  in.  long  drawn 
in  a  northeasterly  direction.  The  line  is  drawn  so  as  to 
represent  the  force  to  some  scale;  here  it  is  10  Ib.  to  the  inch. 
An  arrowhead  is  put  on  one  end  of  the  line  to  show  its  direction. 
In  Fig.  248,  OP  is  a  line  representing  a  directed  quantity. 
Such  a  line  is  called  a  vector.  O  is  the  beginning  of  the  vector 

and  P  is  the  terminal.  OQ 
is  the  projection  of  the  vec- 
tor upon  OX,  and  OR  is  the 
projection  upon  OY.  OQ 
and  OR  are  called  compo- 
nents of  the  vector.  As 
before,  OQ  =  OP  cos  6,  and 

x      OR  =  OP  sin  6. 

Example  I,  Suppose  that 
a  weight  W  is  resting  on  a 
rough  horizontal  table  as  shown  in  Fig.  249.  Suppose  that  a 
force  of  40  Ib.  is  acting  on  the  weight  and  in  the  direction  OP, 
making  an  angle  of  20°  with  the  horizontal;  then  the  horizon- 
tal pull  on  the  weight  is  OQ  =40  cos  20°  =  37.588  Ib.,  and  the 
vertical  lift  on  the  weight  is  OR  =  40  sin  20°  =  13.68  Ib. 

Example  2.  A  car  is  moving  up  an  incline,  making  an  angle 
of  35°  with  the  horizontal,  at  the  rate  of  26  ft.  per  second. 
What  is  its  horizontal  ve- 
locity? Its  vertical  ve- 
locity? 

Solution.  Horizontal  ve- 
locity =  26  cos  35°  =  21 .3  ft. 
per  second.  Vertical  ve- 
locity =  26  sin  35°  =  14.9  ft . 
per  second. 

EXERCISES  117 

1.  The  line  segment  AB  17  in.  long  makes  an  angle  of  33°  with  the 
line  OX.     Find  the  projection  upon  OX.     Find  its  projection  upon  the 
line  OY  perpendicular  to  OX  and  in  the  same  plane  as  OX  and  AB. 

Aiis.  14.258  in.;  9.258  in. 

2.  A  steamer  is  moving  in  a  southeasterly  direction  at  the  rate  of 
24  miles  per  hour.     How  fast  is  it  moving  in  an  easterly  direction?     In 
a  southerly  direction?  Ana.  16.97  mi.  per  hr.  in  each. 


Fio.  249. 


RIGHT  TRIANGLES 


429 


3.  The  eastward  and  northward  components  of  the  velocity  of  a  ship 
are  respectively  5.5  miles  and  10.6  miles.     Find  the  direction  and  the 
rate  at  which  the  ship  is  sailing. 

Ans.   11.94  mi.  per  hr.,  27°  25.4'  east  of  north. 

4.  A  roof  is  inclined  at  an  angle  of  33°  30'.     The  wind  strikes  this 
horizontally  with  a  force  of  1800  pounds.     Find  the  pressure  perpendicu- 
lar to  the  roof.  Ans.  993.4  Ib. 

6.  A  roof  20  ft.  by  25  ft.  and  inclined  at  an  angle  of  27°  25'  with  the 
horizontal  will  shelter  how  large  an  area?  Ans.  443.85  ft.2 

6.  A  hillside  is  on  a  slope  of  16°  and  contains  5.2  acres.     How  much 
more  is  this  than  the  projection  of  the  hillside  upon  a  horizontal  plane? 

Ans.  0.2  acre. 

7.  Show  in  general  that  the  projection  of  a  plane  area  upon  a  fixed 
plane  is  equal  to  the  given  area  times  the  cosine  of  the  angle  between 
the  planes. 

8.  Show  in  general  that  the  component  of  a  force  along  any  fixed  line 
is  equal  to  the  magnitude  of  the  force  times  the  cosine  of  the  angle 
between  the  direction  of  the  force  and  the  fixed  line. 

9.  Two  men  are  lifting  a  stone  by  means  of  ropes.     The  ropes  are  in 
the  same  vertical  plane.     One  man  pulls  85  Ib.  in  a  direction  23°  from 
the  vertical  and  the  other  105  Ib.  in  a  direction  42°  from  the  vertical. 
Determine  the  weight  of  the  stone.  Ans.   156.3  Ib. 


350.  Definitions. — The    angle    of    elevation    is    the    angle 
between  the  line  of  sight  and  the  horizontal  plane  through 


FIG.  250. 


FIG.  251. 


the  eye  when  the  object  observed  is  above  the  horizontal 
plane.  When  the  object  observed  is  below  the  horizontal 
plane,  the  angle  is  called  the  angle  of  depression. 

Tims,  if  0  is  the  object  observed  by  the  eye  at  E,  the  angle 
XEO  is  the  angle  of  elevation  in  Fig.  250,  and  the  angle  of 
depression  in  Fig.  251. 

Directions  on  the  surface  of  the  earth  are  often  given  by 
directions  as  located  on  the  mariner's  compass.  As  seen  from 


430 


PR  A  CTICAL  MA  THEM  A  TICS 


Fig.  252,  these  directions  are  located  with  reference  to  the 
four  cardinal  points,  north,  south,  east,  and  west. 

Directions  are  often  spoken  of  as  bearings. 

When  greater  exactness  is  required,  the  direction  may  lx^ 
given  as  a  certain  number  of  degrees  from  a  cardinal  point. 
Thus,  a  direction  given  north  10°  east  means  a  direction  10° 
oast  of  north;  south  40°  west  means  40°  west  of  south. 


FIG.  252. — Mariner's  compass. 


EXERCISES  118 


1.  If  a  vertical  staff  20  ft.  high  casts  a  shadow  20  ft.  long  on  level 
ground,  find  the  angle  of  elevation  of  the  sun.  Ans.  37°  33.8'. 

2.  How  many  degrees  east  of  north  is  N.E.?     N.N.E.?     N.  by  E.? 
N.E.  by  N.?  Ans.  45°;  22*°;  11}°!  33J°. 

3.  A  flag  staff  70  ft.  high  casts  a  shadow  40  ft.  long.     Find  the  angle  of 
elevation  of  the  sun  above  the  horizon.  Ans.  60°  15.3'. 

4.  At  60  ft.  from  the  base  of  a  fir  tree  the  angle  of  elevation  of  the  top 
is  75°.     Find  the  height  of  the  tree.  Ans.   224  ft.  nearly. 

6.  What  is  the  inclination  from  the  vertical  of  the  face  of  a  wall  having 
a  batter  of  J?  Ans.  7°  7.6'. 

A  batter  of  |  means  that  the  wall  slopes  1  ft.  in  a  rise  of  8  ft. 

6.  What  is  the  angle  of  slope  of  a  road  bed  that  has  a  grade  of  5  per  cent? 
One  with  a  grade  of  0.25  per  cent?  Ans.  2°  51.7';  8.6'. 

7.  Find  the  angle  between  the  rafter  and  the  horizontal  in  the  follow- 
ing pitch  of  roofs:  two-thirds,  half,  third,  fourth. 

Ans.  53°  7.8';  45°;  33°  41.3';  26°  33.9'. 


RIGHT  TRIANGLES 


431 


8.  Certain  lots  in  a  city  are  laid  out  by  lines  perpendicular  to  B  Street 
and  running  through  to  A  Street  as  shown  in  Fig.  253.     Find  the  widths 
of  the  lots  on  A  Street  if  the  angle  between  the  streets  is  28°  40'. 

Ans.  114.0  ft. 

9.  A  man  whose  eyes  are  5  ft.  6  in. 
above  the  ground  stands  on  a  level 
with,   and    150  ft.  distant  from,  the 
foot  of  a  flag  staff  72  ft.  high.     What 
angle  does  his  line  of  sight  when  look- 
ing at  the  top  of  the  staff  make  with 
the  horizontal  line  from  his  eyes  to 
the  pole?  Ans.  23°  54.6'. 

10.  In  surveying  on  the  Lake  Front 
in  Chicago,  measurements  were  taken 

as   shown   in  Fig.    254.     Find    the   distance   on    a   straight   line  from 
A  to  E.  Ans.  338.4  ft. 

11.  In  an  isosceles  triangle  one  of  the  base  angles  is  48°  20',  and  the 


i  r 


FIG.  253. 


C\*~—*>L—*\B 


t— ! 


•114.78-'— 


E 


D 


Coal  Pile 


FIG.  254. 


base  is  18  in.     Find  the  legs,  vertical  angle,  and  the  altitude  drawn  to 
the  base.  Ans.  Legs  13.54  in.;  vert.  ang.  83°  20';  alt.  10.11  in. 

12.  The  side  of  a  regular  pentagon  (five-sided  figure)  is  12  in.     Find 
the  radius  of  the  inscribed  circle,  and  the  area  of  the  pentagon. 

Ans.  Radius  8.258  in.;  area  247.8  in.2 
Suggestion.     Draw   the   pentagon  and 
inscribe  a  circle  as  in  Fig.  255.     Angle 
=  T2°.     Triangle  AOB   is  isosceles. 
6 


We    have    tan  36°  =  - 


-  =  6-r-tan  36° 


angle  CAB -=40°. 
the  length  of  CP. 


Could  this  problem  be  solved  as  easily 
by  geometry? 

13.  Find   a  side   of  the  regular  octa- 
gon  circumscribed   about  a  circle  20  ft. 
in  diameter.  Ans.  8.284  ft. 

14.  Given    the    right    triangle    ABC, 
with  C  the  right  angle,  CB  =  2Q  ft.,  and 

Produce   CB   to    P  making  angle  CAP  =  70°,  find 

Ans.  65.48  ft. 


432 


PR  A  CTICA  L  MA  THE  MA  TICS 


16.  Find  the  shorter  altitude  and  area  of  a  parallelogram  whose  sides 
are  10  ft.  and  25  ft.,  and  the  angle  between  the  sides  75°. 

Ana.  Alt.  9.659  ft. ;  area  241.5  ft.1 

16.  Two  points  C  and  B  are  on  opposite  banks  of  a  river.  A  line  AC 
at  right  angles  to  CB  is  measured  40  rods  long;  the  angle  CAB  is  meas- 
ured and  found  to  be  41°  40'.  Find  the  width  of  the  stream. 

AHS.  35.00  rd. 


17.  Wishing  to  determine  the  width  of  a  river,  I  observed  a  tree 
standing  directly  across  on  the  bank.  The  angle  of  elevation  of  the 
top  of  the  tree  was  32°.  At  150  ft.  back  from  this  point  and  in  the  same 
direction  from  the  tree  the  angle  of  elevation  of  the  top  of  the  tree  was 
21°.  Find  the  width  of  the  river.  Ans.  239  ft.  nearly. 

Suggestion.  Let  a:  =  width  of  river,  and  y  =  height  of  tree.  The  rela- 
tions of  the  parts  are  as  given  in  Fig.  256. 

D        c  (I)tan32°-|- 

B         (2)tan21°  =  ^L, 

Here  are  two  equations  and  two  unknown 
numbers.  The  solution  of  them  will  give 
the  values  of  x  and  y. 

18.  Locate  the  centers  of  the  holes  B  and 
C,  Fig.  257,  by  finding  the  distance  each 
is  to  the  right  and  above  the  center  O.  The 
radius  of  the  circle  is  1.5  in.  Compute  cor- 
rect to  three  decimal  places. 
Ans.  B,  1.2135  in.,  0.8817  in.;  C,  0.4635  in.,  1.4266  in. 

19.  A  man  surveying  a  mine  measures  a  length  AB  =  220  ft.  due  east 
with  a  dip  of  6°  15';  then  a  length  BC=325  ft.  due  south  with  a  dip  of 
10°  45'.     How  much  lower  is  C  than  A1  Ans.  84.57  ft. 

20.  A  building  80  ft.  long  by  60  ft.  wide  has  a  roof  inclined  at  36° 
with  the  horizontal.     Find  the  area  of  the  roof,  and  show  that  the  result 
is  the  same  whether  the  roof  has  a  ridge  or  not.  Ans.  5933  ft.1 

21.  In  the  side  of  a  hill  that  slopes  upward  at  an  angle  of  32°,  a  tunnel 
is  bored  sloping  downward  at  an  angle  of  12°   15'  with  the  horizontal. 
How  far  is  a  point  115  ft.  down  the  tunnel,  below  the  surface  of  the  hill? 

Ans.  94.63  ft. 


Fio.  257. 


RIGHT  TRIANGLES 


433 


22.  The  angle  of  elevation  of  a  balloon  from  a  point  due  south  of  it 
is  60°,  and  from  another  point  1  mile  due  west  of  the  former  the  angle 
of  elevation  is  45°.     Find  the  height  of  the  balloon.     Ans.  1.225  miles. 

23.  From  the  top  of  a  mountain  1050  ft.  high  two  buildings  are  seen 
on  a  level  plane  and  in  a  direct  line  from  the  foot  of  the  mountain.     The 
angle  of  depression  of  the  first  is  35°  and  of  the  second  is  24°.     Find  the 
distance  between  the  buildings.  Ans.  858.8  ft. 


Q 


FIG.  25S. 

24.  If  R  and  r  are  the  radii  of  two  pulleys,  D  the  distance  between  the 
centers,  and  L  the  length  of  the  belt,  show  that  when  the  belt  is  not 
crossed,  Fig.  258,  the  length  is  given  by  the  following  formula  where 
the  angle  is  taken  in  radians: 

R-r 


L=2VD2-(R-r)*+Tr(R+r)+2(R-r)  sin'1 


D 


26.  Show  that  when  the  belt  is  crossed,  Fig.  259,  the  length  is  given  by 
the  following  formula: 


L  = 


R  fr)  2  +  (R  +r)  (*•  +2  sin-1   — 


T 


FKJ.  250. 

Note.  These  formulas  would  seldom  be  used  in  practice.  An  ap- 
proximate formula  would  be  more  convenient,  or  the  length  would  be 
measured  with  a  tape  line. 

A  rule  often  given  for  finding  the  length  of  uncrossed  belts  is:  Add 
twice  the  distance  between  the  centers  of  the  shafts  to  half  the  sum  of 
the  circumferences  of  the  two  pulleys. 

28 


434  PRACTICAL  MATHEMATICS 

26.  In  exercise  24,  given  7?  =  18  in.,  r=8  in.,  and  D  =  12  ft.,  find  the 
length  of  the  belt  by  the  formula.     Find  the  length  by  the  approximate 
rule.  Am.  30.87-  ft.;  30.81-  ft. 

27.  Use  the  same  values  as  given  in  exercise  26,  and  find  by  the  formula 
the  length  of  the  belt  when  crossed.  An*.  31.20—  ft. 

28.  A  belt  connects  two  pulleys  of  diameters  6  ft.  and  2  ft.  respectively. 
If  the  distance  between  their  centers  is  15  ft.,  find  the  length  of  the  belt, 
making  no  allowance  for  slack.  A  ns.  42.83  ft. 

29.  Two  pulleys,  of  diameters  7  ft.  and  2  ft.  respectively,  are  connected 
by  a  crossed  belt.     If  the  centers  of  the  pulleys  are  16  ft.  apart,  find  the 
length  of  the  belt.  An*.  47.41  ft. 

30.  A  chord  of  2  ft.  is  in  a  circle  of  radius  three  feet.     Find  the  length 
of  the  arc  the  chord  subtends  and  the  number  of  degrees  in  it. 

An*.  2.038  ft.;  38°  56.3'. 

Suggestion.  In  Fig.  260,  the  chord  A  B  =  2  f t.  and  the  radius  OA  =  3  ft. 
Triangle  AOC  is  a  right  triangle.  Angle 
AOC  =  \  angle  AOB,  and  the  central  angle 
AOB  has  the  same  measure  as  the  arc  AnB. 

31.  Find  the  area  of  the  sector  AnBO  in  ex- 
ercise  30.     Find   the  area  of  triangle  AOB. 
Find  the  area  of  the  segment  ABn. 

An*.  Sector  3.057  ft.2;  triangle  2.828  ft.1; 
segment  0.229  ft.2 

32.  Find    the    area   of  the  segment  whose 
FIG.  260.                 chord  is  4  ft.  in  a  circle  of  5  ft.  diameter. 

Ans.  2.794  ft.1 

33.  Find  the  area  of  a  segment  whose  chord  is  6  ft.  and  height  2  ft. 

Ans.  8.67  ft.1 

34.  In  a  circle  of  60  in.  radius,  find  the  area  of  a  segment  having  an 
angle  of  63°  15'.     Find  the  length  of  the  chord  and  the  height  of  the  seg- 
ment, take  3  of  their  product,  and  compare  with  the  area  found. 

Ans.  379m.1 

35.  A  cylindrical  tank  resting  in  a  horizontal  position  is  filled  with 
water  to  within  10  in.  of  the  top.     Find  the  number  of  cubic  feet  of  water 
in  the  tank.     The  tank  is  10  ft.  long  and  4  ft.  in  diameter. 

Ans.  106.7  ft.» 

36.  Compute  the  volume  for  each  foot  in  the  depth  of  a  horizontal 
cylindrical  oil  tank  of  length  30  ft.  and  diameter  8  ft. 

Ans.  108.8  ft.3;  294.8  ft.3;  516.4  ft.3;  754  ft.3;  991.6  ft.';  1213.2  ft.*; 
1399.2  ft.3 

Note.  In  the  same  manner,  the  volume  could  be  computed  for,  say, 
each  i  in.  In  this  way  a  gage  could  be  made  for  determining  the  quan- 
tity of  oil  in  a  tank. 

37.  The  slope  of  the  roof  in  Fig.  261  is  30°.     Find  the  angle  0  which  is 
the  inclination  to  the  horizontal  of  the  line  AB,  drawn  in  the  roof  and 
making  an  angle  of  35°  with  the  line  of  greatest  slope. 

An*.  24°  11.1'. 


RIGHT  TRIANGLES 


435 


38.  Find  the  angle  between  the  diagonal  of  a  cube  and  one  of  the  diag- 
onals of  a  face  which  meets  it.  Ans.  3-5°  15.9'. 

39.  A  hill  has  a  slope  of  32°.     A  path  leads  up  it  making  an  angle  of 
45°  with  the  line  of  greatest  slope.     Find  the  slope  of  the  path. 

Ana.  22°  0.3'. 

40.  Two  set  squares,  whose  sides  are  3,  4,  and  5  in.,  are  placed  as  in 
Fig.  262  so  that  their  4-in.  sides  and  right  angles  coincide,  and  the  angle 
between  the  3-in.  sides  is  50°. 

Find   the  angle  0  between  the 
longest  sides.      Ans.  29°  22.5'. 

41.  What  size  target  at  30  ft, 
from  the  eye  subtends  the  same 
angle   as   a  target  4  ft.  in  di- 
ameter at  1000  yd.?     Find  the 
angle  it  subtends. 

Ans.  0.48  in. ;  4.6'.  FIG.  261. 

42.  The  description  in  a  deed 

runs  as  follows:  "Beginning  at  a  stone,  A,  at  the  N.W.  corner  of  lot 
401;  thence  east  112  ft.  to  a  stone,  B;  thence  S.  36|°  W.  100ft.; 
thence  west  parallel  with  A  B  to  the  west  line  of  said  lot  401 ;  thence  north 
on  west  line  of  said  lot  to  the  place  of  beginning."  Find  the  area  of  the 
land  described.  Ans.  6612.88  ft.2 

43.  If  the  point  of  observation  is  at  a  distance  of  h  feet  above  the 
surface  of  the  earth,  find  the  farthest  distance  that  can  be  seen  on  the 
surface  of  the  earth;  that  is,  find  the  distance  of  the  horizon. 


FIG.  262. 


FIG.  263. 


Discussion.  In  Fig.  263,  let  0  be  the  center  of  the  earth,  r  the  radius 
of  the  earth,  and  h  the  height  of  the  point  P  above  the  surface;  it  is 
required  to  find  the  distance  from  the  point  P  to  the  horizon  at  A. 


PRACTICAL  MATHEMATICS 

(PA)1  -  (PQ)'-(QA)'  -  (r+A)«-r»  -2rA+/i». 


436 


For  points  above  the  surface  that  are  reached  by  man,  h*  ia  very  small 
as  compared  with  2rh. 

•'•  PA  =  \/2rh,  approximately. 

Here  PA,  r,  and  h  are  in  the  same  units.     Now  let  h  be  in  feet,  and 
r  and  PA  be  in  miles.     Also  let  r  =  3960  miles.     Then 


PA 


=  -y/ 


2  X  3960  X 


~ 


miles. 


We  may  then  state  the  following  approximate  rules: 
The  distance  of  the  horizon  in  miles  is  approximately  equal  to  the  square 
root  of  $  times  the  height  of  the  point  of  observation  in  feet. 


The  height  of  the  point  of  observation  in  feet  is  j>  times  the  square  of  the 
distance  of  the  horizon  in  miles. 

Definition.     The  angle  APC  =  6  is  called  the  dip  of  the  horizon. 

44.  Find  the  greatest  distance  at  which  the  lamp  of  a  lighthouse  can 
be  seen  from  the  deck  of  a  ship.     The  lamp  is  85  ft.  above  the  surface 
of  the  water  and  the  deck  of  the  ship  30  ft.  above  the  surface. 

Ana.  18  mi.  approx. 

45.  A  cliff  2000  ft.  high  is  on  the  sea  shore ;  how  far  away  is  the  horizon  ? 
What  is  the  dip  of  the  horizon?  Ans.  54.8  mi.  approx.;  47'. 


Fio.  265. — Acme  thread. 

46.  Find  the  radius  of  a  circle  circumscribed  about  a  polygon  of  128 
sides  if  one  side  is  2  in.  What  is  the  difference  between  the  circumference 
of  the  circle  and  the  perimeter  of  the  polygon? 

Ans.  40.81  in.;  0.417  in. 


RIGHT  TRIANGLES 


437 


47.  In  Whitworth's  English  Standard  screw  threads,   Fig.   264,  the 
angle  between  the  sides  of  the  threads  is  55°.     If  the  top  of  a  thread  is 
rounded  off  |  of  the  height  and  the  bottom  filled  in  the  same  amount, 
find  the  depth  to  four  decimal  places  of  the  threads  of  the  following 
pitches:  1,  8,  14,  and  26. 

Ans.  0.6403  in.;  0.0800  in.;  0.0457  in.;  0.0246  in. 

48.  In  an  acme  thread,  the  angle  between  the  sides  of  the  threads  is 
29°.     When  the  pitch  is  P,  the  depth  and  the  other  dimensions  are  as 
shown  in  Figs.  265,  267.     (a)  Suppose  that  the  top  dimensions  and  the 
depth  are  given,   find  the  dimensions  at  the  bottom,     (b)    Find  the 
dimensions  at  the  top  and  bottom  and  the  depth  for  an  8-pitch  acme 
thread. 

Ans.  (b)  0.0787  in.;  0.0463  in.;  0.0793  in.;  0.0457  in.;  0.0638  in. 

49.  In  a  worm  thread,  the  angle  between  the  sides  is  29°.     The  dimen- 
sions are  as  shown  in  Figs.  266,  267.      (a)  Suppose  the  top  dimensions 


14  Vi 


FIG.  266. — Worm  thread. 

and  the  depth  are  given,  find  the  bottom  dimensions,  (b)  Find  the 
dimensions  in  a  7-pitch  thread.  (The  above  are  the  Brown  and  Sharp 
proportions.)  What  are  the  differences  between  a  worm  thread  and  an 
acme  thread?  Ans.  (b)  0.0950  in.;  0.0479  in.;  0.0443  in.;  0.0981  in. 

60.  From  a  point  A  on  a  level  with  the  base  of  a  steeple  the  angle  of 
elevation  of  the  top  of  the  steeple  is  42°  30';  from  a  point  B  22  ft.  directly 
over  A  the  angle  of  elevation  of  the  top  is  36°  45'.  Find  the  height  of 
the  steeple  and  the  distance  of  its  base  from  A. 

Ans.  118.9  ft.;  129.8  ft. 


Worm  Thread 


Acme  Thread 
FIG.  267. 


U.S.S.  Thread 


61.  A  ship  sailing  due  north  observes  two  lighthouses  in  a  line  due 
west ;  after  an  hour's  sailing,  the  bearings  of  the  lighthouses  are  observed 
to  be  southwest  and  south-southwest.  If  the  distance  between  the 
lighthouses  is  8  miles,  at  what  rate  is  the  ship  sailing? 

Ans.  13.66  mi.  per  hour. 


438 


PRACTICAL  MATHEMATICS 


50 
62.  Show  that  R  =  - — v-7y    where   R  =  the   radius  of  curvature  and 

D  =*  the  degree  of  the  curve. 

361.  Widening  of  pavements  on  curves. — The  tendency  of 
a  motorist  to  "cut  the  corners"  is  due  to  his  unconscious  desire 
to  give  the  path  of  his  car  around  a  turn  the  longest  possible 
radius.  Many  highway  engineers  recognize  this  tendency  by 
widening  the  pavement  on  the  inside  of  the  curve  as  shown  in 
Fig.  268.  The  practice  adds  much  to  the  attractive  appear- 
ance of  the  highway.  If  the  pavement  is  the  same  width 
around  the  curve  as  on  the  tangents,  the  curved  section  appears 
narrower  than  the  normal  width ;  whereas  if  the  curved  section 
is  widened  gradually  to  the  mid-point  G  of  the  turn,  the  pave- 
ment appears  to  have  a  uniform  width  all  the  way  around. 

In  order  that  the  part 
added  may  fit  the  curve 
properly  it  is  necessary  to 
have  the  curve  of  the  inner 
edge  a  true  arc  of  a  circle, 
tangent  to  the  edge  of  the 
straightaway  sections,  and 
therefore  it  must  start 
before  the  point  E  of  the 
curve  is  reached.  The 
part  added  may  be  easily 
staked  out  on  the  ground  with  transit  and  tape,  by  means 
of  data  derived  from  the  radius  r,  the  central  angle  6  of  the 
curve,  and  the  width  w.  In  practice  the  width  w  is  taken 
from  2  ft.  to  8  ft.  according  to  the  value  of  r.  The  width  added 
can  be  readily  computed  when  values  for  r,  w,  and  8  are  given. 
Referring  to  the  figure,  derive  the  following  formulas: 


x  —  r  sec  $0— r  = 


cos 


.  —  r. 


x-\-w  =  r'  sec  \d  —  r' = 
.'.  r'  = 


cos 

x+w        (x-\-w)  cos 


sec  $0  —  1        1  — cos  $0 
t  =  r  tan  $0. 
'  =  r'  tan  i0. 


RIGHT  TRIANGLES 


439 


Area  added  =BFCEAG  =  BPAO'-FPEC-BGAO'. 
BPAO'  =  r'tr. 

FPEC  =  FPEO  -FCEO  =  rt  - 


360 


.'.  area  added  =r't'-    « 


*       - 


=  r't'-rt- 


6 

360 


7r(r'+r)(r'  — r). 


Exercise.  Find  the  number  of  square  feet  in  the  area  added  if  r  =300 
ft.,  w=4  ft.,  and  0  =  100°.  Ans.  1395  ft.2 

352.  Spirals. — If  a  line  is  drawn  around  a  circular  cylinder  so 
that  it  advances  a  certain  distance  along  the  cylinder  for  each 
revolution,  the  curve  thus  formed  is  a  spiral  or  a  helix. 

If  a  piece  of  paper  is  cut  as  shown  in  (a)  Fig.  269,  and  lines 
AB  and  CD  drawn,  this  piece  of  paper  can  be  rolled  into  the 


D 


B 


D 


B,C 


(a) 


FIG.  269. 


cylinder  (6)  Fig.  269,  where  the  lines  AB  and  CD  of  (a)  form 
the  spiral  running  from  A  to  D  of  (6). 

The  advance  along  the  cylinder  for  each  turn  of  the  spiral 
is  the  lead  of  the  spiral,  or  the  spiral  lead.  In  Fig.  269, 
AC  is  the  lead.  It  is  customary  to  give  the  lead  of  the  spiral 
as  so  many  inches  per  one  turn.  For  instance,  a  spiral  that 
advances  8  in.  in  one  turn  is  called  an  8-in.  spiral. 


440  PRACTICAL  MATHEMATICS 

The  angle  a  that  the  spiral  makes  with  an  element  of  the 
cylinder  is  the  angle  of  the  spiral.     It  is  seen  that 

_  circumference  of  cylinder 

lead  of  the  spiral 
C'R 
or,  in  (a)  Fig.  269,  tan  a  .= 


In  setting  milling  machines  for  cutting  spirals  such  as  worms, 
spiral  gears,  counter  bores,  and  twist  drills,  it  is  often  necessary 
to  know  the  angle  of  the  spiral. 

To  find  the  angle  of  a  spiral  or  for  the  cutters  in  cutting  a 
spiral,  make  a  drawing  as  shown  in  Fig.  270;  the  angle  C 
being  a  right  angle,  CB  the  circumference,  and  AC  the  lead. 
Angle  A  is  the  angle  required,  and  may  be  measured  with  a 

fin 

protractor,  or  it  may  be  found  by  finding  tan  A  =  -j~  and  using 


the  table  of  tangents. 

For  ready  reference  the  following  rules  are  given: 


Lead 


FlO.   270. 

ANGLE.  Divide  the  circumference  of  the  spiral  by  the  lead 
(advance  to  one  turn) ,  and  the  quotient  is  the  tangent  of  the  angle 
of  the  spiral. 

LEAD.  Divide  the  circumference  of  the  spiral  by  the  tangent 
of  the  angle,  and  the  quotient  is  the  lead  of  the  spiral. 

CIRCUMFERENCE.  Multiply  the  tangent  of  the  angle  by  the 
lead  of  the  spiral,  and  the  product  will  be  the  circumference. 

When  applying  calculations  to  spiral  gears,  the  angle  is 
reckoned  at  the  pitch  circumference. 

EXERCISES  119 

1.  Find  the  angle  of  the  spirals  in  the  following  twist  drills: 

(1)  Diameter  of  drill  &  in.,  lead  2.92  in.  Ans.  18°  35.2'. 

(2)  Diameter  of  drill  1  i  in.,  lead  9.33  in.  Am.  20°  44.8'. 

(3)  Diameter  of  drill  tf  in.,  lead  7.29  in.  Ans.   19C  13.3'. 


RIGHT  TRIANGLES 


441 


2.  Find  the  angle  of  the  spiral  thread  on  a  double-threaded  worm  of 
pitch  diameter  3|  in.  and  having  three  threads  in  2  in.    Ans.  83°  5.1'. 

3.  In  a  tower  the  outer  diameter  of  a  winding  stairway  is  12  ft.     Find 
the  angle  of  the  spiral  formed  by  the  outer  end  of  the  steps  if  the  stairway 
makes  one  turn  in  ascending  18  ft.     Find  the  angle  of  the  spiral  formed 
by  the  inner  ends  of  the  steps  if  the  steps  are  4  ft.  long. 

Ans.  64°  28.7';  34°  55.1'. 

4.  The  piece  shown  in  Fig.  271  has  a  length  of  85  in.  and  a  diameter 
of  |  in.     If  the  spiral  grooves  make  a  half  turn  in  the  length  of  the  piece, 
find  the  angle  of  the  spiral.  Ans.  15°  40'. 

6.  Find  the  angle  for  setting  cutters  in  cutting  the  following  spirals: 
(a)  Diameter  jin.  and  lead  2.78  in.;  (b)  f  in.  and  7.62  in.;  (c)  2  in.  and 
10,37  in.;  (d)  1  in.  and  22.5  in. 

Ans.  (a)  15°  46.7';  (b)  17°  10.9';  (c)  31°  13.1';  (d)  6°  59.2'. 


FIG.  271. 

6.  A  cylinder  2  in.  in  diameter  is  to  have  spiral  grooves  making  angles 
of  20°  with  the  center  line  of  the  cylinder.     What  will  be  the  lead  of  the 
spiral?  Ans.  17.26  in. 

7.  Find  the  length  of  one  turn  of  a  spiral  around  a  cylinder  3  in.  in 
diameter  if  the  lead  of  the  spiral  is  9  in.  Ans.  13.03  in. 

8.  Show  that  the  length  of  any  spiral  is  given  by  formula 

L=n\/C2+/2, 

where  L=  length  of  spiral,  n=the  number  of  turns  of  the  spiral,  C  =  cir- 
cumference of  cylinder  the  spiral  is  on,  and  Z=lead  of  the  spiral. 

9.  Find  the  length  of  a  spiral  making  20  turns  in  8  in.  on  a  cylinder 
3.5  in.  in  diameter.  Ans.  220.0  in. 


CHAPTER  XXXIX 

RELATIONS  BETWEEN  RATIOS,  AND 
PLOTTING 

353.  Relations  between  the  ratios  of  an  angle  and  the  ratios 
of  its  complement — In  Art.  327  it  was  pointed  out  that  the 
function  of  an  acute  angle  is  equal  to  the  co-function  of  its 
complement.     This  gives  the  following  relations.     It  can  easily 
be  shown  that  these  relations  hold  for  any  value  of  the  angle  6. 

[62]  sin  (90° -6)=  cos  6. 
[63]  cos  (90° -6)=  sin  6. 
[64]  tan  (90°  -6)=  cot  6. 
[66]  cot  (90°  -6)=  tan  0. 
[66]  sec  (90° -8)=  esc  6. 
[67]  esc  (90° -6)=  sec  6. 

As  previously  explained,  it  is  these  relations  that  make  a 
table  of  trigonometric  functions  do  double  duty. 
Example,     sin  60°  =  sin  (90°  -  30°)  =  cos  30°. 

354.  Relations  between  the  ratios  of  an  angle  and  the  ratios 
of  its  supplement. — The  following  relations  between  the  ratios 
of  an  angle  and  its  supplement  are  true  for  any  value  of  0. 
They  are  convenient  when  one  wishes  to  find  by  means  of  the 
tables  the  functions  of  angles  lying  between  90°  and  180°. 

[68]  sin  (180° -6)=  sin  6. 
[69]  cos  (180°-6)  =  -cos  6. 
[70]  tan  (180° -6)  = -tan  6. 
[71]  cot  (180° -6)  = -cot  6. 
[72]  sec  (180°-6)  =  -sec  6. 
[73]  esc  (180°-e)=csc  6. 

Proof.  In  Fig.  272,  angle  XOP  =  6  is  any  angle,  and  angle 
XOQ  =  18Q°-8. 

From  any  point  in  the  terminal  side  of  XOP,  as  B,  draw  the 
perpendicular  AB  to  the  z-axis;  and  from  D,  any  point  in  the 

442 


RELATIONS  BETWEEN  RATIOS,  AND  PLOTTING     443 

terminal  side  of  XOQ,  draw  the  perpendicular  CD  to  the  z-axis. 
The  right  triangles  OAB  and  OCD  are  similar.  Also  OA,  AB, 
OB,  CD,  and  OD  are  positive,  while  OC  is  negative. 


Then  sin  (180°  -  0)  =  gg  = 

' 

AJso   cos 


=  sn 


Examplel.  sin  130°  15'  =  sin  (180°-49°  45')  =  sin  49°  45' 
=  0.7633. 

Example  2.  sos  160°  =  cos  (180°  -  20°)  =  -cos  20°  = 
-0.9397. 


FIG.  272. 


355.  Relations  between  ratios  of  an  angle  6  and  90° +6. — 

The  following  relations  connect  the  ratios  of  any  angle  6 
and  90° +  0.  They  are  also  convenient  to  use  in  finding,  by 
means  of  the  tables,  the  functions  of  angles  lying  between 
90°  and  180°. 

[74]  sin  (90° +6)=  cos  0. 
[75]  cos  (90° +6)  = -sin  0. 
[76]  tan  (90° +6)  = -cot  6. 
[77]  cot  (90°+e)  =  -tan  6. 
[78]  sec  (90°+6)  =  -csc  6. 
[79]  esc  (90° +0)=  sec  6. 

Proof.     In    Fig.    273,    angle   XOP  =  8   and   angle   XOQ  = 

90°+ e. 

From  any  point  in  the  terminal  side  of  each  draw  a  perpen- 


Ill 


PRACTICAL  MATHEMATICS- 


•  I  in  ilar  to  the  z-axis.     The  right  triangles  AOB  and  OCD  thus 
formed  are  similar,  and  have  all  their  sides  positive  except  OC. 

Then  sin  (90°+0)  =  ^  =  ~  =  cos  d. 

(JL)      (Jo 


Also  tan  (90°+0)  == 


=  -cot  d, 


Example  1.     sin  130°  15'=sin  (90°+40°  15')=  cos  40°  15' 
=  0.7633. 

Example  2.     tan  116°  20'  =  tan  (90°+26°  20')  =  -cot  26°  20' 
=  -2.0204. 


Fio.  273. 


366.  Relations  between  the  ratios  of  an  angle  and  the 
ratios  of  its  negative. — The  following  relations  connect  the 
ratios  of  positive  and  negative  angles,  and  are  true  for  any 
value  of  0. 

[80]  sin   (-0)  =  -sin  6. 
[81]  cos  (-6)  =cos  6. 
[82]  tan  (-6)  =  -tan  6. 
[83]  cot  (-6)  = -cot  6. 
[84]  sec  (-0)  =  sec  0. 
[86]  esc  (-6)  =  —esc  tt. 

Proof.     In  Fig.  274,  angle  XOP  =  0,  and  angle  XOQ=-d. 
From  any  point  in  the  terminal  side  of  each  draw  a  per- 
pendicular to  the  x-axis.     The  right  triangles  OA B  and  OAC 


RELATIONS  BETWEEN  RATIOS,  AND  PLOTTING      445 

thus  formed  are  similar,  and  have  all  their  sides  positive  but 
AC,  which  is  negative. 


ml  •  /  n\  ^r  At) 

Then  sin  (  —  0)  =  -^  =  —  -=-=  =  —  sm  6. 
UL          Uo 

.     M     OA        OA 
And  cos  (  —  0)=7T7;  =      pr^  =  cos0. 
t/C  UJD 

Also  COt     (  —  6)=  TTr  —  —  rn  =  ~  c°t  &' 

AC/          AD 

Example  1.     sin  (-30°)  =  -sin  30°=  -0.5. 

Example  2.     tan  (  -  46°  10')  =  -  tan  46°  10'  =  -  1  .0416. 


EXERCISES  120 

Draw  a  figure  in  each  case  and  prove  the  following: 

1.  sin  (90° -0)  =cos  6.  7.  esc  (180°  -0)  =csc.  0. 

2.  tan  (90° -0)  =cot  0.  8.  cos  (90° +  0)  =  -sin  0. 

3.  sec  (90°  -0)  =csc  0.  9.  cot  (90° +  0)  =  -tan  0. 

4.  tan  (180°  -0)  =  -tan  0.  10.  sec  (90°  +  0)  =  -esc  0. 
6.  cot  (180° -0)  =  -cot  0.                11.  tan(-0)  =  -tan  0. 

6.  sec  (180°-0)  =  -sec  0.  12.  sec  (-0)  =sec.  0. 

By  means  of  the  tables  find  the  following: 

13.  sin  140°.  16.  cos  (-49°).  19.  sin  159°  40'. 

14.  cos  150°.  17.  tan  (-17°).  20.  cos  (-117°  30')- 

15.  tan  170°.  18.  cot  (-125°).  21.  tan  (-156°  10'). 
Ans.  Ex.  14.   -0.8660.     Ex.  17.   -0.3057.     Ex.  20.  -0.4617. 


357.  Relations  between  the  ratios  of  any  angle. — As  stated 
in  Art.  324, 

[86]  sin  8  = or  esc  0  =  - — -• 

esc  0  sin  6 

[87]  cos  0  = or  sec  6  = -. 

sec  6  cos  6 

[88]  tan  0  ==  — —  or  cot  6  = 


cot  6  tan  6 

By  definition, 

[89]  versedsine  0,  abbreviated  vers  6  =  1  — cos  6. 

[90]  coversedsine  0,  abbreviated  covers  0  =  1  —  sin  6. 
The  truth  of  the  following  can  easily  be  proved  when  0 
is  any  angle. 


446 


PRACTICAL  MATHEMATICS 

[91]  sin'O  +  cos2  6  =  1. 
[92]  sec26  =  l+tan2e. 
[93]  csc20  =  l+cot26. 


[96,  cot  .- 
Proo/  o/  [91].     By  definition  (see  Fig.  275), 

7/2  X2 

sin2  0  =  —„  and  cos2  5  =  -= 
r2  r2 

Hence  adding,  sin2  0+cos2  8  =  ^+^  =  V      *  • 

But  t/2-|-x*  =  r2  because  the  triangle  OCZ?  is  a  right  triangle. 


FIG.  275. 

r2 
Hence  sin2  0+cos2  0  =  —  =  1. 

r2 

Proo/  of  [92].     By  definition, 

sec2  0  =  —  ,  and  tan2  0  =  —  • 
x2  x2 

r2  ?/2 

/.  sec2  6  =  1+  tan2  B  because  -,  =  1  +  ^-  = 

X2  X2 

EXERCISES  121 


1.  Prove  [93],  [94],  and  [96]  by  using  the  definitions. 

2.  From  [91]  derive  (a)  sin  6  =  ±\/l—  cos*  0,  (b)  cos  9  —  ± 
8.  Solve  [92]  for  sec  0  and  tan  0.    __ 

An*,  sec  0=  ±-v/l+tan2  0;  tan  0—  ±-v/sec*  0  —  1. 
4.  Solve  [93]  for  esc  6  and  cot  6.  __ 

Ans.  esc  0=  ±\/l-r-cot8  0;  cot  0=  f-y/csc1  0-1. 


RELATIONS  BETWEEN  RATIOS,  AND  PLOTTING     447 

5.  From  the  relations  already  given,  write  five  other  relations  between 
the  ratios. 

6.  Prove  the  following  relations : 

(1)  cos  6  tan  0=sin  6.  (3)   sec  6  cot  0=csc  6. 

(2)  sin  0  cot  8  =  cos  6.  (4)  tan  6  esc  8  =  sec  0. 
Use  the  formulas  and  find  the  other  functions : 

7.  When  sin  0  =  %. 

8.  When  cos  0  =  f . 

9.  When  tan  0=3. 

10.  Find  the  following  from  Table  XI :  (1)  sin  122U  20',  (2)  cos  110°  44', 
(3)  tan  163°  15',  (4)  cot  172°  50',  (5)  cos  (-42°  16'),  (6)  tan  (-21°  49'). 
Ans.  (1)  0.8450;  (2)   -0.3540;  (3)   -0.3010;  (4)   -7.9530;  (5;  0.7400; 
(6)  -0.4003. 

368.  Plotting    the    sine    curve. — In  Chapter    XXXIII   the 

plotting  of  the  curve  of  an  algebraic  equation  was  considered. 
The  trigonometric  equations  can  be  represented  by  curves 
plotted  in  a  similar  manner. 

y 


30°  15° 60°     90°    120°135'l50\lSO°    210°     240°    270°    300°     330°     360 


FIG.  276. 

Example.     Plot  the  curve  for  y  =  sin  x. 

Choose  suitable  angles  for  values  of  x  and  take  the  values 
for  y  from  the  table  of  Art.  332  or  from  Table  XI. 

Values  of  x,  0°  30°     45°     60°  90°    120°    135°    150°    180°. 

Values  of  y,  0  0.5      0.7    0.87      1    0.87      0.7      0.5        0. 

Values  of  x,      210°  225°    240°  270°  300°   315°     330°   360°. 

Values  of  y,  -0.5  -0.7  -0.87    -1-0.87-0.7-0.5     0. 

Choose  a  convenient  unit  on  the  z-axis  and  y-axis  and  plot 
the  points  as  shown  in  Fig.  276,  using  the  angle  as  abscissa  and 
the  sine  as  ordinate. 

The  curve  in  Fig.  276  is  called  the  sine  curve.     It  extends 


•its 


PRACTICAL  MATHEMATICS 
Y 


Fio.  277. 


3r 


y  =  tan  z- 


Fio.  278. 


Fio.  279. 


RELATIONS  BETWEEN  RATIOS,  AND  PLOTTING     449 

both  ways  indefinitely,  repeating  the  part  given.  That  is, 
if  values  between  360°  and  720°  be  taken  the  curve  will  be 
found  to  be  of  the  same  shape.  The  same  will  be  true  for 
values  of  the  angle  less  than  0°. 

A  curve  that  repeats  itself  in  this  way  is  called  a  periodic 
curve.  The  equation  giving  rise  to  it  is  called  a  periodic 
function. 

The  relative  lengths  of  the  units  representing  x  and  y  may 
be  changed,  and  the  curve  thus  drawn  out  or  shortened. 
However,  the  curve  will  still  be  of  the  same  general  form. 

359.  Curves  for   cosine,   tangent,   cotangent,   secant,  and 
cosecant. — By  choosing  suitable  angles  for  values  of  x  and 
determining  the  corresponding  values  of  y,  the  coordinates  of 
points  are  obtained  which,  when  plotted,  give  the  curves  of 
Figs.  277,  278,  and  279. 

EXERCISES  122 

1.  Plot  y  =  sin  x,  using  various  units  for  x  and  y. 

2.  Plot  y=cos  x  from  x  =  0°  to  x  =  360°  and  get  form  of  Fig.  277. 

3.  Plot  y=tan  x  and  y=cot  x  from  x=0°  to  z  =  360°  and  get  forms 
of  Fig.  278. 

4.  Plot  y  =sin  z+cos  x  from  x  =  0°  to  x  — 180°. 

5.  Plot  y  =cos2  x—  sin2  x  from  x  =  Q  to  £  =  |?r. 

00 

6.  Plot  7/=cos-1T  from  x=0  to  a;  =  3. 

7.  Plot  y  =sec  x  and  y  =  csc  x  from  x  =0°  to  x  =  360°  and  get  the  forms 
of  Fig.  279. 

360.  Projections  of  a  point  having  circular  motion. — 

Example  1.  A  point  P,  Fig.  280,  moves  around  a  vertical 
circle  of  radius  3  in.  in  a  counter-clockwise  direction.  It 
starts  with  the  radius  in  a  horizontal  position  and  moves  with 
an  angular  velocity  of  one  revolution  in  10  seconds.  Plot 
a  curve  showing  the  distance  from  the  center  0  of  the  projec- 
tion of  P  on  the  vertical  diameter  at  any  time. 

Discussion.  Describe  a  circle  with  center  0  and  radius 
3  in.,  to  scale  if  desired.  In  this  circle  the  point  P  starts  at  A. 
After  1  second  OP  has  turned  to  the  position  of  OPi  through  an 
angle  of  36°  =  0.6283  radians,  after  2  seconds  to  the  position 
OP  %  through  an  angle  of  72°  =  1.2566  radians,  and  so  on  to  the 
positions  OP5,  OP4,  •  •  •  OP10. 

29 


1.-.0 


PR  A  CTICAL  MA  THEM  A  TICS 


The  points  NI,  Ni,  etc.,  are  the  projections  of  PI,  P8,  etc., 
on  the  vertical  diameter. 

Produce  the  horizontal  diameter  OA,  and  lay  off  the  seconds 
on  this  to  some  scale. 

For  each  second  plot  a  point  whose  ordinate  is  the  corre- 
sponding distance  of  TV  from  0.  Fractions  of  seconds  may 
be  taken  and  more  points  located  if  desired.  Connect  these 
points,  and  we  have  a  curve  for  which  any  ordinate  y  is  the 
distance  from  the  center  O  of  the  projection  of  P  on  the  vertical 
diameter  at  the  time  t,  represented  by  the  abscissa  of  the  point. 

It  is  to  be  noted  that  the  ordinate  of  the  curve  increases 
from  0  to  +3,  decreases  from  +3  to  0  and  from  0  to  —3, 


f 

•" 

.   j 

D 

r 

\ 

P 

^ 

s 

P    / 

N 

2 

/ 

\ 

/ 

Ai 

// 

1\ 

V 

i  

\ 

A 

\ 

h 

\ 

_  Y 

\ 

0 

^ 

A   ; 

'lice 

iu  S< 

'. 

cond 

' 

s 

\  I 

* 

j 

i 

/ 

V? 

E 

,/ 

\ 

/ 

* 

X 

X 

\ 

^-ji 

/ 

P 

P 

1 

FIG.  280. 

and  increases  from  —3  to  0.  The  curve  is  made  up  of  two 
like-shaped  parts,  one  below  and  one  above  the  horizontal  line. 
Together  they  form  a  cycle  of  the  curve. 

It  is  evident  that,  for  the  second  and  each  successive  revo- 
lution, the  curve  repeats  itself,  that  is,  is  a  periodic  curve. 

Since  OP  turns  through  36°  per  second,  the  angle  AOP  = 
36<°  =  0.6283J  radians.  Because  of  the  right  triangle  OPN, 

y=ON  =  OP  sin  36J°  =  3  sin  36<°. 

Or,  in  general,  y—r  sin  wt,  where  o>  (Greek  letter  omega) 
is  the  angle  turned  through  in  one  second,  that  is,  the  angular 
velocity.  It  is  then  the  equation  of  the  curve. 

It  is  readily  seen  that  if  a  straight  line  of  length  r,  such  as 
a  crank,  starts  in  a  horizontal  position  when  t  =  0,  and  revolves 
in  a  vertical  plane  around  one  end  at  a  uniform  angular  velocity 


RELATIONS  BETWEEN  RATIOS,  AND  PLOTTING     451 


u>  per  second,  the  projection  of  the  moving  end  on  a  vertical 
straight  line  has  a  motion  which  is  represented  by  the  equation 

y  =  r  sin  coi. 

Similarly,  the  projection  of  the  moving  end  on  a  horizontal 
straight  line  is  given  by  the  ordinates  of  the  curve  whose 
equation  is 

y  =  r  cos  o)t. 

If  OP  represents  the  crank  of  an  engine  with  a  connecting 
rod  which  is  very  long  compared  with  OP,  the  motion  of  the 
cross-head  is  represented  approximately  by  that  of  the  point 
.V,  and,  hence,  by  the  equation  y  =  r  sin  cot. 

A  simple  periodic  motion  like  the  above  is  sometimes  repre- 
sented by  the  equation  y  =  r  sin  cot  and  sometimes  by  y  =  r 
Y 


P, 


'45    .5 


FIG.  281. 

cos  ut.  The  first,  if  the  time  is  measured  from  the  instant 
when  OP  is  in  the  line  of  the  simple  periodic  motion,  and  second, 
if  measured  from  the  instant  when  OP  is  perpendicular  to  the 
line  of  the  simple  periodic  motion. 

If  the  time  is  counted  from  some  other  instant  than  the 
above,  we  have  y  =  r  sin  (o^+a),  where  a  is  the  angle  that  OP 
makes  with  the  line  OA  at  the  instant  t  is  counted  from. 

Example  2.  A  crank  OP,  Fig.  281,  of  length  2  ft.  starts 
from  a  position  making  an  angle  a  =  40°  with  the  horizontal 
line  OA  when  t  =  0.  It  rotates  in  a  vertical  plane  in  the  posi- 
tive direction  at  the  rate  of  2  revolutions  per  second.  Plot  a 
curve  showing  the  projection  of  P  on  a  vertical  diameter. 

Discussion.  As  before,  draw  a  circle  of  2-ft.  radius  to  some 
scale.  Extend  the  horizontal  axis  to  the  right,  and  represent 
the  part  of  a  second  necessary  for  one  revolution  to  some 
suitable  scale. 


452  PRACTICAL  MATHEMATICS 

When  <  =  0,  OP  starts  from  the  position  OPo  making  an 
angle  .4OP0  =  a  =  40°.  OP  turns  through  720°  in  1  sec.,  or 
36°  =  0.0283  radians  in  0.05  sec.  Here  we  plot  a  point  for  every 
0.05  sec.  Any  other  convenient  fraction  of  a  second  could 
be  chosen.  The  position  of  the  free  end  of  the  crank  for  each 
0.05  sec.  until  a  complete  revolution  is  made  is  located  by 
Po,  PI,  P2,  PS  •  •  •  .  These  are  at  intervals  of  36°  starting 
with  40°. 

At  any  time  t,  the  position  of  OP  makes  with  OA  an  angle 
AOP=o)t+a.  And  the  distance  from  O  to  the  projection  of 
P  at  N  is 

ON  =  OP  sin  (w*+a), 
or  O#  =2  sin  (720<°+a). 

The  curve  of  this  example  is  periodic.  It  repeats  itself 
for  each  complete  revolution  of  the  crank. 

It  follows  that,  in  general,  the  equation  of  the  curve  is 
2/=rsin  (ut+a). 

361.  Sine  curves  of  different  frequency.  —  The  frequency 
refers  to  the  number  of  cycles  of  the  periodic  curve  in  a  unit 
of  time.  Thus,  if  a  crank  makes  4  revolutions  in  1  second, 
the  curve  showing  the  motion  makes  4  cycles  in  1  second; 
while  the  curve  for  a  crank  that  makes  2  revolutions  in  1  second 
makes  2  cycles  in  1  second.  The  frequency  of  the  first  is 
4,  of  the  second  2. 

The  angular  velocity  w  divided  by  360°  when  measured  in  de- 
grees, or  by  2ir  when  measured  in  radians,  gives  the  frequency  ft 


The  time  necessary  for  one  cycle  is  called  the  period.     It 
is  evident  that  the  period  T  is  given  by  the  formula 


Consider  the  following  equations: 

Equation         Angle  at    Angular    Frequency  Period 

velocity  u              u  r  _  ?T 

/~2r  ^~w 

(1)  y  =  r  sin}/              \t            4              0.0796  12.566 

(2)  y  =  r  sin*                   t             1               0.1592  6.283 

(3)  y  =  r  sin2t              2t            2              0.3183  3.1416 


RELATIONS  BETWEEN  RATIOS,  AND  PLOTTING     453 


If  r  is  taken  as  1  and  the  three  curves  plotted,  they  are  as 
in  Fig.  282. 

It  is  to  be  noted  that  the  frequency  of  a  sine  curve  varies  directly 


y  =  Bin  y2  t 

=8in 
1J   =  sin  2  t 


-1  - 


FIG.  282. 


as  the  value  of  u,  that  is,  as  the  coefficient  of  t  in  the  angle  ut. 
The  number  a>  is  sometimes  called  the  frequency  factor. 
362.  Variation  in  the  amplitude  of  sine  curves. — The  am- 


0 


-1 


—2 


/    \  y*=  h  sin  e  - 

I             \  n 

\  y=  sin  a       — 

•.  y  =  2  sin  0    - 

\  W=  2  sin  3^- 


\ 


\J 


y  i 

\./  --.....x  v 


FIG.  283. 


plitude  of  a  sine  curve  is  the  difference  between  its  greatest 
and  least  ordinate. 

In  the  previous  article  all  of  the  curves  plotted  have  the 


454  PRACTICAL  MATHEMATICS 

same  amplitude,  and  it  will  be  remembered  that  the  radius 
in  each  case  was  taken  equal  to  unity. 
Consider  the  curves  of  the  equations: 

(1)  y  =  \  sin  0,  (3)  y  =  2  sin  0, 

(2)  2/  =  sin  0,  (4)  y  =  2  sin  36. 
These  curves  are  plotted  in  Fig.  2&3. 

It  is  to  be  noted  that  the  amplitude  of  a  sine  curve  varies  directly 
as  the  coefficient  of  sin  6,  that  is,  as  the  value  of  r  in  the  generating 
circle. 

EXERCISES  123 

1.  A  crank  18  in.  long  starts  in  a  horizontal  position  and  turns  in  a 
positive  direction  in  a  vertical  plane  at  the  rate  of  0.7854  radian  per 
second.     The  projection  of  the  moving  end  of  the  crank  upon  a  vertical 
line  oscillates  with  a  simple  periodic  motion.     Construct  a  curve  whose 
ordinates  show  the  distance  of  the  projection  from  the  center  of  its  path 
at  any  time. 

2.  Write  the  equation  of  the  curve  in  the  previous  exercise.     Find  the 
value  of  the  ordinate  when  t  =0.5.     When  t  =2.3. 

Ans.  y  =  18  sin(0.7854<);  6.89  in.;  17.50  in. 

3.  A  crank  8  in.  long  starts  in  a  position  making  an  angle  of  55°  with 
the  horizontal,  and  rotates  in  a  positive  direction  at  the  rate  of  20  revolu- 
tions per  minute.     Draw  a  curve  to  show  the  projection  of  the  moving 
end  of  the  crank  on  a  vertical  line. 

4.  Write  the  equation  of  the  curve  in  the  last  exercise.     Find  the 
value  of  y  when  t  =  1.5  seconds. 

Ans.  t/  =  8  sin  (120<°+55°);  -6.55  in. 
Plot  the  curves  that  represent  the  following  motions: 

5.  y  =3.5  sin  (35*°  +36°). 

6.  y  =  12sin  (1.88*  +0.44). 


8.  Plot  y=r  sin  -£  and  i/  =  r  sin  (-<+  -Ion   the   same   axes.     Notice 

that  the  highest  points  are  separated  by  the  constant  angle  -       Such 

4 

curves  are  said  to  be  out  of  phase.     The  difference  in  phase  is  stated  in 
time  or  as  an  angle;  in  the  latter  case  it  is  called  the  phase  angle. 

9.  Plot  (1)  y=r  sin  ^t,   (2)  y  =  r  sin  (f-^)  '  and  (3)  y  =r  cos  ^t,  all 
on  the  same  axes.     What  is  the  difference  in  phase  between  (1)  and  (2)? 

10.  Plot  on  the  same  set  of  axes:  (1)   y=r  sin  rt,    (2)  y=r  sin  ^t, 

(3)  y-rsin  ^(. 

11.  Plot  on   the  same  set  of  axes:   (1)  //=40ain  9,  (2)  »/=30sin  9, 
(3)  y=20sin  29. 


CHAPTER  XL 


TRIGONOMETRIC  RATIOS  OF  MORE  THAN 
ONE  ANGLE 

363.  In  the  present  chapter  will  be  given  a  number  of  formu- 
las of  very  great  use  in  more  advanced  subjects  of  mathematics 
and  in  their  applications.     These 

formulas  are  given  for  reference. 
But  little  attempt  will  be  made 
here  to  prove  them  or  illustrate 
their  uses. 

364.  Functions  of  the  sum  or 
difference  of  two  angles.  —  The 
following    formulas   express   the 
functions  of  the  sum  or  the  dif- 
ference of  two  angles  in  terms  of 
the    functions    of    the    separate 
angles.      These     formulas     are 

sometimes  called  the  addition   and  subtraction  formulas  of 
trigonometry. 

[96]  sin   (A+B)=sin  A  cos  B+cos  A  sin  B. 

[97]  sin   (A—  B)=sin  A  cos  B-cos  A  sin  B. 

[98]  cos  (A+B)=cos  A  cos  B  —  sin  A  sin  B. 

[99]  cos  (A-B)=cos  A  cos  B+sin  A  sin  B. 

MAAI  A      (K  \-o\      tan  A  +  tan  B 
[100]  tan 


FIG.  284. 


tan  A  —  tan  B 


Proof  of  [96]  for  A<90°,  £<90°,  and  (A+5)<90°.  In 
Fig.  284,  angle  XOQ  =  A  and  angle  QOP  =  B.  Hence  angle 
XOP  =  A+B. 

Draw  NP,  QP,  QR,  and  MQ  perpendicular  respectively  to 
OX,  OQ,  NP,  and  OX.  Then  angle  RPQ  =  A. 

NP    MQ+RPMQ    RP 


455 


450  PRACTICAL  MATHEMATICS 

But  MQ  =  OQ  sin  A,  and  RP  =  QP  cos  A,  from  the  right 
triangles.     Putting  these  values  in  place  of  MQ  and  RP, 

OQ  sin  A  .  QP  cos  A 


.".  sin  (A+Z?)  =  sin  A  cos  B-j-cos  A  sin  /?. 
Formulas  [97]  to  [101]  may  be  proved  in  a  similar  manner. 
Example  1.     Given  the  functions  of  45°  and  30°,  find  the 
value  of  sin  15°. 

Solution,     sin  15°  =  sin  (45°  -30°) 

=  sin  45°  cos  30°  -cos  45°  sin  30° 
=  0.7071X0.8660-0.7071X0.5=0.2588. 
Example  2.     Test    formula    [100]    by    using   A  =  16°    and 
B=27°. 

Substituting  these  values  in  the  formula, 

tan  a6°+27°}-  Jan_160+tan_270 
~l-tanl6°tan276' 

Putting  in  the  values  of  tan  16°  and  tan  27°  from  Table  XI, 

0.2867+0.5095 
=  r^o7286Txo:5095  =  a9325- 

366.  Functions  of  twice  an  angle  and  half  an  angle.  —  By 
putting  B  =  A  in  the  formulas  for  the  sum  in  Art.  364,  the 
following  formulas  are  obtained: 

[102]  sin  2A     2  sin  A  cos  A. 

[103]  cos  2A  =  cos2  A  -sin2  A  =  1-2  sin2  A  =  2  cos2  A  -  1. 

,104]  taa  2A  = 


These  formulas  are  used  to  express  the  function  of  any  angle 
in  terms  of  the  functions  of  one-half  that  angle. 

From  the  above  may  be  derived  the  following  formulas, 
which  are  used  to  express  the  functions  of  any  angle  in  terms 
of  the  functions  of  twice  that  angle. 


MAKI  ,*-cos  2A 

[106]  sm  A=  H 


Mnei  /1-f  COS  2A 

[106]  cos  A-+-J 

\         & 

MA_,  /I  — cos  2A        sin  2A        1  — cos  2A 

[107J  tan  A=  ±\/1 


'l-f-cos2A     1+cos  2A        sin  2A 


TRIGONOMETRIC  RA  TIOS  OF  MORE  THAN  ONE  ANGLE     457 

Example  1.     Test  formula  [102]  by   using  A  =20°. 
Substituting  in  the  formula, 

sin 40°  =  2  sin 20°  cos 20°  =  2X0.3420X0.9397  =0.6428. 
Example  2.     Test  formula  [106]  by  using  A  =  35°. 
Substituting  in  the  formula, 

/I  +  cos  70°        /I +0.3420 
cos  35  =  A/ ~ =A/ ^ —     =0.81915. 

366.  Formulas  for  changing  products  td  sums  or  differ- 
ences, and  sums  and  differences  to  products. — The  following 
four  formulas  are  convenient  for  expressing  a  product  of  two 
functions  as  the  sum  or  difference  of  two  functions. 

[108]  sin  A  cosB  =  |  sin  (A+B)+|  sin  (A-B). 
[109]  cos  A  sin  B  =  |  sin  (A+B)-£  sin  (A-B). 
[110]  cosAcosB  =  |  cos  (A+B) +  ^  cos  (A-B). 
[Ill]  sin  A  sin  B=  -|  cos  (A+B)  +  £  cos  (A-B). 

The  following  four  formulas  express  a  sum  or  a  difference 
as  a  product.  They  are  often  convenient  when  working  with 
logarithms. 

[112]  sin  A+sin  B  =  2  sin  |  (A+B)  cos  |  (A-B). 
[113]  sin  A-sin  B  =  2  cos  \  (A+B)  sin  \  (A-B). 
[114]  cosA+cosB  =  2  cos£  (A+B)  cos  HA— B). 
[115]  cos  A — cos  B  =  -2  sin  \  (A+B)  sin  \  (A-B). 

EXERCISES  124 

1.  Find  sin  90°,  cos  90°,  and  tan  90°,  using  the  addition  formulas  and 
90°  =  60° +30°. 

2.  Find  sin  (A  +B)  if  sin  A  =  •  and  sin  B  =  jj,  both  being  acute  angles. 
Find  cos  (A+B).  Ans.  1;  0. 

3.  Find  cos    (A—  B)  if  cos  A  =  A  and  cos  5={|,  both  being  acute 
angles.     Find  sin  (A—  B).  Ans.  iff;  iJ|. 

4.  Find  tan  (A  +B)  if  tan  A  =2  and  tan  B  =3,  both  being  acute  angles. 
What  is  the  value  of  (.4  +5)?  Ans.  - 1 ;  135°. 

6.  Given  the  functions  of  30°  and  45°,  find  sin  75°,  cos  75°,  and  tan  75°. 

6.  Take  values  from  Table  XI  for  the  functions  of  25°  and  18°,  and  find 
sin  43°,  cos  43°,  sin  7°,  and  tan  7°. 

7.  Take  values  from  Table  XI  for  the  functions  of  15°  and  find  sin  30°, 
cos  30°,  and  tan  30°. 

8.  Using  the  functions  of  45°,  find  sin  22£°,  cos  22£°,  and  tan  22^°- 

9.  Given  sin  20  =  £,  find  tan  0  and  cot  6.  Ans.   ±£;  ±3. 


458  PRACTICAL  MATHEMATICS 

10.  Show  that  COB  50  +cos  30  =  2  cos  40  cos  0. 

11.  Show  that  sin  70—  sin  30-2  cos  50  sin  20. 

12.  Show  that  cos  125°  -cos  75°  =  -2  sin  100°  sin  26°. 

13.  Express  sin  7o+sin  5a  as  a  product.  Ana.  2  sin  6a  cos  a. 

14.  Express  cos  46°+cos  28°  as  a  product.       Ann.  2  cos  37°  cos  9° 

15.  Show  that  sin  575°  cos  927°+cos  575°  sin  927°  -sin  1602°. 

.    tan327°+tan_846° 

16.  Show  that  ^^3270^8^0  =  tan  1173  . 

.     -TT    j  Ai_  i  .     .      11*  15*-  9r  OT 

17.  Innd  the  value  of  sin  ——  cos  ---  htan  —r  cot  -75-- 

o  4  4  i 

Ann.  -0.3536. 

18.  Find  the  value  of 


ll«-\   .     19r  .          /     4»\   .     /     7r\ 

~  T  )  8m  T  +  C08  (~  37  8m  \  ~"37  ' 


19.  If  sin  a  =  —  $  and  a  is  in  III  quadrant,  and  cos  0  =  }  and  ft  is  in  1 
quadrant,  find  the  value  of  sin  (a+0),  sin  (a—  0),  and  tan  (a+0). 

Ans.  -0.7618;  0.5475;  1.1760. 

.  sin  47°+sin  17° 

20.  Show  that ._0         -r=5=tan  32°. 

cos  47  +cos  17 

,,    .  sin  3«+sin  5a 

21.  Show  that  —  —  =cota. 

cos  3a  —cos  5a 

,,    ,  sin  67°— sin  23° 

22.  Show  that  C0867o+cos23o=tan  22°. 

23.  In  a  right  triangle  cos  a—  -  and  cos  /3  =  -  •     Show  that 

c  c 

]c  —  b       .    .     i-       ^  Ic  —  a. 

and  sin  i/3  =  \/~o~~ 

24.,If  a  is  less  than  360°,  in  what  quadrants  may  a  be  if  sin  i  a  is  nega 
tive?  Positive?  If  tan  \a  is  negative?  Positive?  If  cot  $a  is  nega 
tive?  Positive?  If  sec  i  a  is  negative?  Positive? 


CHAPTER  XLT 
SOLUTION  OF  OBLIQUE  TRIANGLES 

367.  Cases. — A  triangle  that  is  not  a  right  triangle  is  called 
an  oblique  triangle. 

The  student  should  now  reread  Arts.  338  and  339  and  the 
exercises  following  them.  It  is  evident,  then,  that  an  oblique 
triangle  can  be  solved  when  there  are  three  elements  given  at 
least  one  of  which  is  a  side. 

There  arise  the  four  following  cases: 

CASE   I.     When  any  side  and  any  two  angles  are  given. 

CASE  II.  When  any  two  sides  and  the  angle  opposite  one 
of  them  are  given. 


CASE  III.  When  any  two  sides  and  the  angle  included 
between  them  are  given. 

CASE  IV.     When  the  three  sides  are  given. 

The  oblique  triangle  can  be  divided  into  right  triangles 
by  drawing  convenient  altitudes,  and  so  be  solved  by  methods 
already  given  in  the  chapter  on  right  triangles.  It  is,  however, 
usually  a  saving  of  time  to  solve  them  by  means  of  formulas 
derived  especially  for  that  purpose.  The  simpler  of  these 
formulas  will  now  be  derived. 

368.  The  law  of  sines. — The  law  is:  In  any  triangle,  the 
sides  are  proportional  to  the  sines  of  the  opposite  angles. 

Proof.  Let  ABC  in  Fig.  285  be  any  triangle.  Draw  the 
altitude  h  from  B  to  the  side  A  C.  Because  of  the  right  triangles, 
in  either  (a)  or  (6),  h  —  c  sin  A  and  h  =  a  sin  C, 

459 


400 


PR  A  CTICAL  MA  THEM  A  TICS 


.'.  a  sin  C  =  c  sin  A. 
Dividing  both  members  of  this  by  sin  C  sin  A, 

a       c_ 

sin  A    sin  C 

Similarly  by  drawing  an  altitude  from  C,  - — 

'  sm  A     sin  B 

b  c 


[116]  .'. 


a 


tions: 


sin  A     sin  B     sin  C 

From  the  law  of  sines  there  may  be  written  the  three  equa- 

a  b  a  c  b  c 

sin  A     sin  B'   sin  A     sin  C'   sin  B    sin  (f 
any  one  of  which  involves  four  elements  of  a  triangle.     It  is 
evident  that  if  any  three  of  the  elements  involved  in  one  of 
the  equations  are  given,  the  remaining  element  can  be  found. 

Thus,  if  in  - — -r  =  - — 5'  A,  B,  and  6  are  given,  then  solving 
sin  A     sm  B 


for  a,  a  = 


sin 
b  sin  A 
sin  B 


369.  The  law  of  cosines. — The  law  is:  In  any  triangle,  the 
square  of  any  side  equals  the  sum  of  the  squares  of  the  other  two 
sides  minus  twice,  the  product  of  these  two  sides  times  the  cosine 
of  the  angle  between  them. 


O) 


b  D 


C  A 


¥m.  2JSC. 


Proof.     In  Fig.  280,  either   (a)   or   (6),   draw   (ho  altitude 
h  from  B  to  the  side  AC.     Let  AD  =  m  and  DC  =  n. 

Because  of  the  right  triangles,  in  either  (a)  or  (6),  a2  =  A2+n2. 

In  (a),  n  =  b— m  =  b  —  c  cos  A. 

In  (b),  n  =  m  —  b  =  c  cos  A— 6. 

In  either  case,  n2  =  62— 26c  cos  A-f-c2  cos2  A. 

In  either  (a)  or  (6),  h*  =  c*  sin2  A. 


SOLUTION  OF  OBLIQUE  TRIANGLES 


461 


Substituting  these  values  for  h2  and  n'2  in,  a2  = 

a2  =  c2  sin2  A+b2—2bc  cos  A+c2  cos2  A. 
:.  a2  =  &2+c2  (sin2  A+cos2  A)-2bc  cos  A. 
[117]  .'.  a2  =  b2+c2-2bc  cos  A. 
Similarly  there  may  be  obtained: 

[118]  b2=a2+c2-2accosB. 
[119]  c2  =  a2+b2-2abcosC. 
Solving  these  for  cos  A,  cos  B,  and  cos  C  respectively: 


[120]  cos  A  = 
[1211  cos  B  = 
[122]  cos  C  = 


b2+c 


2_a2 


2bc 


2ac 

a2+b2-c2 
2ab 


By  these  formulas  the  values  of  the  angles  of  a  triangle  can 
be  computed  when  the  sides  are  known. 

370.  Directions  for  solving. — It  will  be  noticed  that  each 
of  the  formulas  from  the  law  of  sines  and  the  law  of  cosines 
involves  four  elements  of  the  triangle. 

In  solving  a  triangle,  select  a  formula  that  involves  three 
known  elements  besides  the  element  that  is  to  be  found,  solve  for 
the  unknown  element  in  terms  of  the  known,  substitute  the  numer- 
ical values,  and  evaluate. 

The  work  may  be  checked  (a)  by  making  a  careful  construc- 
tion, and  (b)  by  using  some  other  formula  than  is  used  in  solving. 

371.  Case  I,  a  side  and  two  angles  given. — Example.     Given 
a -45,  5  =  36°  17',  and  C  =  83°  32';  to  find  6,  c,  and  A. 


Formulas. 
(1)  A=--180°-(B+C). 


Construction. 


(2) 


sin  A     sin  B' 


/.&  = 


sin  A     sin  C' ' 


a  sin  B 

sin  A 

a  sin  C 

sin  A 


Computation  by  natural  functions. 
A  =  180° -(36°  17'+83°  32')=60C 
a  sin  B    45 X. 5918 


11' 


c  — 


sin  A 
a  sin  C 
sin  A 


.8676 

45  X. 9937 

.8676 


=  30.70. 


=  51.54. 


FIG.  287. 


462 


1>K  A  CT1CA  L  M  A  Til  KM  A  TICS 


Computation  by  logarithms. 

log  45=    1.6532 
Iogsin36°17'  =  9.7722 

11.4254 
log  sin  60°  11'=  9.9384 

log  6  =    1.4870 
.  6  =  30.69 


log  45=    1.6532 
log  sin  83°  32'=  9.9972 

11.6504 
log  sin  60°  11'=   9.9384 


log  c=    1.7120 
/.  c  =  51.53 


Check  by  using  the  law  of  cosines. 


372.  Case  II,  two  sides  and  an  angle  opposite  one  of  them 
given. — With  these  parts  given  it  is  possible  (a)  that  there  is 
only  one  solution,  that  is,  that  only  one  triangle  can  be  found; 
(b)  that  there  are  two  solutions,  that  is,  that  there  are  two 
different  triangles  that  fulfill  the  conditions;  (c)  that  there  is 
no  solution,  that  is,  that  no  triangle  exists  that  will  fulfill  the 
conditions. 

Whether  there  is  one  solution,  two  solutions,  or  no  solution 
can  readily  be  determined  by  making  a  careful  construction 
of  the  triangle  from  the  given  parts. 

Example  1.  Given  a  =  15,  c=10,  and  A  =40°  30';  to  find 
6,  B,  and  C. 

The  construction  shows  that  there  is  only  one  solution. 


Formulas. 


Construction. 


.'.  sin  C  = 


sin  A     sin  C 
c  sin  A 


(2)  £  =  180°-(A+C). 

a  b  a  sin  B 


Fio.  288. 


sin  A     sin  B' 


sin  A 


Example  2.  Given  a  =  20,  c  =  25,  and  A  =52°  40';  to  find 
6,  B,  and  C. 

From  the  construction  it  is  seen  that  there  are  two  triangles, 
ABC  and  ABC',  that  fulfill  the  conditions. 


(1) 


SOLUTION  OF  OBLIQUE  TRIANGLES 

Formulas.  Construction, 

a  c 


463 


.'.  sin  C  = 


c  sin  A 


sin  A     sin  C 

(2)  C"  =  180°-C. 

(3)  5  =  180°-(A+C). 

(4)  B'=ABC' 

a  b  7     a  sin  B 


a 


(5) 
(6) 


sin  A     sin  J5 


sin  A     sin  5' 


.'.6'  = 


sin  A 
a  sin  5 
sin  A 


Example  3.     Given  a  =  12,  c = 20,  and 
A  =62°  20';  to  find  6,  B,  and  C.  FK}  28g 

From  the  construction,  Fig.  290,  it  is  evident  that  the  side 
a  is  not  long  enough  to  form  a  triangle.  Hence  there  is  no 
solution. 


FIG.  290. 

373.  Case  III,  two  sides  and  the  angle  between  them 
given.  —  Example.  Given  6=45.2,  a  =  56.7,  and  C=47°  45'; 
to  find  c,  A,  and  B. 

Construction. 
B 


Formulas. 

(1)  c=V«2+62-2a6  cos  C. 
C 


(2)  = 

sin 

.'.  sm  A= 


sin  A     sin 
a  sin  C 


.'.  sin  B  — 


sin  5     sin  C 
b  sin  C 


464 


PRACTICAL  MA  THEM  A  TICS 


Computation. 
c  =  V56.72+45.22-2X56.7X45.2X0.6723  -  42.56. 


n      . 


=80°  26'. 


C/iecfc.     A+5+C  =  180°. 

80°  26'  +51°  49'+47°  45'  =  180°. 

Logarithms  cannot  be  used  conveniently  with  formulas  such 
as  (1)  above.  Formulas  in  which  logarithms  can  be  used 
for  the  solution  of  examples  tinder  Case  III  can  be  derived 
from  the  law  of  sines.  They  are  numbers  123,  124,  and  125 
on  page  473. 

374.  Case  IV,  three  sides  given.  —  Example.  Given  a  =10, 
6  =  12,  and  c=15;  to  find  A,  B,  and  C. 


(1)  cos  A  = 

(2)  cos  B  = 

(3)  cosC  = 


Formulas. 

62+c2-a2 


Construction. 


2bc 

a2+c2-b* 
2oc 


2ab 

Computation. 
122+152-102 
-2XT2X-15-  =  0-7472' 

R     102+152-122 
cos  B  =  — ^TTTTTT-r^r-  =  0.6033, 


cos  C  = 


2X10X15 
102+122-152 


=  0.0792, 


Fm.  292. 

A  =41°  39'. 
#  =  52°  53.5'. 
C  =  85°  27.6'. 


2X10X12 

Check.     ^+B+C  =  180°. 

41°  39'  +  52°  53.5'+85°  27.6'  =  180°  0.1'. 

The  above  formulas  are  not  suitable  for  work  with  loga- 
rithms; but  formulas  can  be  derived  from  them  that  lend 
themselves  readily  to  logarithmic  work.  These  formulas 
are  given  on  page  473,  numbers  126  to  134  inclusive. 


SOLUTION  OF  OBLIQUE  TRIANGLES 


465 


EXERCISES  126 

1.  From  the  law  of  sines,  find  6  in  terms  of  a,  A,  and  B.     Find  C  in 
terms  of  6,  c,  and  B. 


.        ^_g  sin  B .  n  _  .   _t  c  sin  B 
Ans.  u —     ;      j j  C  — sin  r      • 


sin  A 

With  the  following  three  elements  given,  find  those  remaining  and 
check  the  results  when  the  answers  are  not  given. 

2.  A  =44°  6.5';  5  =  57°  42.5';  a  =4.23  in. 

3.  A  =48°  39.2';  C  =  115°  23.8';  a  =  14.83  in. 

Ans.  5  =  15°  57';  6  =5.428  in.;  c  =  17.85  in. 

4.  5=30°  36.8';  C  =  107°  15.5';  6  =  144. 
6.  C  =  44°  17.3';  6  =  14.33;  c  =  13.67. 

Ans.  5  =  47°  2.5';  A  =88°  40.2';  a  =  19.57;  £'  =  132°  57.5';  A'=2°  45.2'; 
a' =0.940. 

6.  A  =53°  16.5';  c  =  25.64;  a  =31.4. 

7.  a  =  79.8;  6=46.7;  5  =  23°  19.6'. 

8.  6  =  17;c  =  16;  A  =47°  16.4'. 

9.  a  =99.4;  c  =  90.4;  B  =  11°  7.8'. 

Ans.  A  =110°  20.4';  C  =  58°  31.8';  6  =  20.5. 

10.  a  =  21;  6=24;  c  =  31. 

11.  a  =  61.52;  6  =  81.74;  c  =  75.34. 

Ans.  A  =45°  53.4';  5  =  72°  33.2';  C  =  61°  33.4'. 

12.  a  =  2.46;  6=3.5;  c  =  4.2. 

13.  To  find  the  distance  AB  through  the  swamp,  Fig.  293,  the  following 
data  were  measured:  a  =  120  rd.,  6  =  146  rd.,  and  C  =  41°  25'.     Compute 
the  distance  AB.  Ans.  97.14  rd. 

14.  Compute  the  inaccessible  distance  AB,  Fig.  294,  from  the  measured 
data  6  =450  ft.,  A  =82°  30',  and  C  =  67°  42'.  Ans.  837.2  ft. 

15.  Two  points,  P  and  Q,  Fig.  295,  are  on  opposite  sides  of  a  stream 
and  invisible  from  each  other  on  account  of  an  island  in  the  stream.     A 
straight  line  AB  Is  run  through  Q,  and  the  following  measurements 
taken:    AQ  =  75Q  ft.,   QB  =  562  ft.,   angle  £AP=47°  28.6',  and  angle 

'  =  57°  45'.     Compute  QP.  Ans.  852  ft. 


FIG.  293. 


FIG.  294. 


16.  From  a  point  on  a  horizontal  plane  the  angle  of  elevation  of  the  top 
of  a  hill  is  23°  46' ;  and  a  tower  45  ft.  high  standing  on  the  top  of  the  hill 
subtends  an  angle  of  5°  16'.     Find  the  height  of  the  hill.      Ans.  173  ft. 
30 


466 


PRACTICAL  MATHEMATICS 


17.  Two  observers,  A  and  B,  100  rd.  apart  on  a  horizontal  plane  ob- 
serve at  the  same  instant  an  aviator.  His  angle  of  elevation  at  A  is 
68°  25  and  at  B  55°  58.2'.  The  angles  in  the  horizontal  plane  made 
by  the  projections  of  the  lines  of  sight  with  the  line  AB  are  43°  27'  at  A 
and  23°  45'  at  B.  Find  the  height  of  the  aviator.  Ans.  1820  ft. 

18.  B  is   42  miles  from  A  in  a  direction  W. 
22°  N.,  and  C  is  58  miles  from  A  in  a  direction  E. 
73°  N.     What  is  the  position  of  C  relative  to  Bl 
Ana.  68.6  mi.  E.  35°  21.1'  N. 

376.  Resultant  of  forces. — If  two  forces, 
represented  by  the  vectors  PQ  and  PS, 
Fig.  296,  act  upon  a  body  at  point  P,  then 
the  combined  effect  of  these  forces  is  the 
same  as  that  of  the  force,  represented  by 
the  vector  PR,  where  PR  is  the  diagonal 
of  the  parallelogram  of  which  PQ  and  PS 
Fio.  295.  are  two  sides. 

The   force  PR  is  called   the  s  R 

resultant    of     the    forces    PQ 
and  PS. 

The  resultant  of  any  number 
of  forces  is  a  single  force  that 
will  produce  the  same  effect  as 
the  combined  effect  of  all  the 
given  forces. 


FIG.  290. 


The  resultant  of  any  number  of 
given  forces  can  bd  found  by  find- 
ing the  resultant  of  any  two  of  the 
given  forces,  then  the  resultant  of 
a  third  force  and  the  first  resultant, 
continuing  till  all  the  forces  are 
used. 

Thus,  if  a,  6,  c,  and  d,  Fig.  297, 
are  four  forces  acting  at  the  point 
P,  r\  is  the  resultant  of  a  and  b,  r2 
the  resultant  of  r\  and  c,  and  r3  the 
resultant  of  r2  and  d.     Therefore  r3 
is  the  resultant  of  a,  b,  c,  and  d. 
376.  Computation   of  a  resultant. — If  two   forces   act   at 
right  angles  to  each  other,  their  resultant  is  evidently  equal 


Fio.  297. 


SOLUTION  OF  OBLIQUE  TRIANGLES 


467 


in  magnitude  to  the  square  root  of  the  sum  of  the  squares  of 
their  respective  magnitudes. 

Thus,  in  Fig.  298,  r=\/az-\-b2.     Also  the  direction  cxf  r 

can  be  found,  for  tan  QPR  =  — 


If  two  forces  act  at  any  angle  0  to  each  other,  their  resultant 
can  be  found  by  using  the  law  of  cosines,  Art  369,  and  is  equal 
in  magnitude  to  the  square  root  of  the  sum  of  the  squares  of  their 
respective  magnitudes  increased  by  twice  their  product  times 
the  cosine  of  the  angle  between  the  two  forces. 

Thus,  in  Fig.  299,  r  =  \/a2-}-b2  —  2ab  cos  <p  by  the  cosine  law. 
But  <p  =  180°  -  6,  and  therefore  cos  <p  =  cos  (180° -0)  =  -cosflby 
[69].  Substituting  this  value  for  cos  <f>,  r  =  \/a2+&2-4-2a6  cos  0. 

The  angle  between  the  resultant  and  either  force  can  be 
found  by  the  law  of  sines,  Art.  368. 

Velocities  can  be  combined  in  exactly  the  same  manner  as 
forces. 

EXERCISES  126 

1.  Given  two  forces  of  40  Ib.  and  60  lb.,  acting  at  an  angle  of  30°. 
Find  the  magnitude  of  their  resultant.  Ans.  67.66+  lb. 

2.  Given  a  force  of  500  lb.   acting  toward  the  east  and  a  force  of 
700  lb.  acting  northeast.     Find  the  resultant  in  magnitude  and  direction. 

Ans.  A  force  of  1111.3  lb.  acting  26°  26.9'  N.  of  E. 

3.  An  aeroplane,  which  is  at  an  altitude  of  1600  ft.  and  moving  at  the 
rate  of  100  miles  per  hour  in  a  direction  due  east,  drops  a  bomb.     Dis- 
regarding the  resistance  of  the  air,  where  will  the  bomb  strike  the  ground 
if  during  its  fall  it  is  acted  upon  by  a  wind  of  40  miles  an  hour  from  a 
direction  30°  east  of  south? 

.4ns.  1279  ft.  23°  24'  N.  of  E.  of  point  where  bomb  was  dropped. 
Solution.     To  find  the  number  of  seconds  it  is  in  falling,  use  equation 
of  exercise  30,  page  339,  \gP  =  1600,  where  g  =  32.     This  gives  t  =  10,  the 
time  in  seconds. 


468  PRACTICAL  MATHEMATICS 

It  would  move  east  from  the  starting  point  as  far  as  the  aeroplane 
travels  in  10  seconds  if  the  wind  is  not  considered. 
100X5280X10 


60X60 

During  10  seconds  the  wind  would  carry  the  bomb 
40X5280X10 


60X60 
The  resultant  of  these  displacements  is  _ 

V/14671+586.7*+2Xl467X586.7Xcos    120°  =  1279  ft 
By  the  sine  law  the  direction  is  found  to  be  23°  24'  N.  of  E. 

4.  An  automobile  is  traveling  W.  36°  N.  at  27  miles  per  hour,  and  the 
wind  is  blowing  from  the  N.  E.  at  30  miles  per  hour.     What  velocity 
and  direction  does  the  wind  appear  to  have  to  the  chauffeur? 

Ana.  37.09  mi.  per  hr.  from  N.  59'  W. 

5.  A  train  is  running  at  30  miles  per  hour  in  a  direction  W.  35°  S., 
and  the  engine  leaves  a  steam  track  in  the  direction  E.  10°  N.     The 
wind  is  known  to  be  blowing  from  the  N.  E.  ;  find  its  velocity. 

Ans.  22  mi.  per  hr. 

6.  In  a  river  flowing  due  south  at  the  rate  of  4  miles  per  hour,  a  boat 
is  drifted  by  a  wind  blowing  from  the  southwest  at  the  rate  of  15  miles 
per  hour.     Determine  the  position  of  the  boat  after  40  minutes  if  resist- 
ance reduces  the  effect  of  the  wind  70  per  cent. 

Ans.  2.  19  mi.  E.14°  26'  S. 


TABLES 

I.  SUMMARY  OF  FORMULAS. 
II.  USEFUL  NUMBERS. 

III.  DECIMAL  AND  FRACTIONAL  PARTS  OF  AN  INCH. 

IV.  ENGLISH  INCHES  INTO  MILLIMETERS. 

V.  U.  S.  STANDARD  AND  SHARP  V-THREADS. 
VI.  CHORDS  OF  ANGLES  IN  CIRCLES  OF  RADIUS  UNITY. 
VII.  STANDARD  GAGES  FOR  WIRE  AND  SHEET  METALS. 
VIII.  SPECIFIC  GRAVITIES  AND  WEIGHTS  OF  SUBSTANCES. 
IX.  STRENGTH  OF  MATERIALS. 
X.  FOUR-PLACE  TABLE  OF  LOGARITHMS. 
XI.  TABLE  OF  NATURAL  AND  LOGARITHMIC  SINES,  COSINES, 
TANGENTS,  AND  COTANGENTS  OF  ANGLES  DIFFERING  BY 
TEN  MINUTES. 


469 


TABLE  I 
Summary  of  Formulas 


11] 

A 

=  ab,  rectangle,  parallelogram. 

[2] 

a 

=  A-5-6,  rectangle,  parallelogram. 

[3] 

b 

=  A-j-a,  rectangle,  parallelogram. 

[4] 

A 

=  $ab,  triangle. 

[5] 

a 

=-2A  -j-b,  triangle. 

[6] 

b 

=  2A-5-a,  triangle. 

[7] 

A 

=  \/s(8—  a)(s—  b)(s  —  c),  where  «  =  i(a+b+c),  triangle. 

[8] 

A 

=  JCB+b)  Xa,  trapezoid. 

[9] 

c=\/a2+62,  right  triangle. 

[10] 

a 

=VC2—  b2,  right  triangle. 

[11] 

b 

=  \/c2  —  a2,  right  triangle. 

[12] 

I 

:  1  =t  :  T,  tapers. 

D-d^L 

[13] 

X 

"~      rt     ^  7  J  tflpcrs. 

£t                  I 

[14] 

D, 

1.732     ,         _     , 

=  D        \r   '  snarp  V-tnreads. 

1  2<W 

[15] 

D1 

—  n       ^'^^^     TT    (3     0    tV,rf>oHo 

—  u         -..j    t   u.  o.  >3.  inn'iicis. 

[16] 

r 

(iw)2+A2  n              _f  _.  ,_ 

—       ^yT  )  segment  ot  circle. 

[17] 

h 

=r—  \/r2  —  (Jtc)2,  segment  of  circle. 

[18] 

w  =  2\/h(2r  —  h),  segment  of  circle. 

[19] 

C 

=rd,  circle. 

[20] 

d 

=  C-:-5r,  circle. 

[21] 

C 

=  2nr,  circle. 

[22] 

2r 

=  C-f-ir,  circle. 

[23] 

A 

=  ^Cr,  circle. 

[24] 

A 

=  nT2,  circle. 

[25] 

A 

=  \-ird*  =  0.7854rf2,  circle. 

[26] 

r 

=  \/A-J-T,  circle. 

[27] 

d 

=  \/A  -j-  JTT,  circle. 

[28] 

Ar 

=  A-a=*R*-irr*  =  ir(Rl-r*)=*(R+r)(R-r),  ring. 

[29] 

A 

=  g^X»r*,  sector. 

I30(a)] 

A 

/t1 
=  §Aw+o—  »  segment. 

I30(b)] 

A 

=  JA*  Vy-  0.608,  segment. 

131] 

A 

=  Tab,  ellipse. 

[32(a)J 

P 

—  T(a-f-b),  ellipse. 

470 


TABLES  471 


[32(c)] 

P=7r\/2(a2+b2),  ellipse. 

[33] 

S  =  ph,  prism. 

[34] 

T  =  ph+2A,  prism. 

[35] 

r  =  6o2,  cube. 

[36] 

p=S  +  h,  prism. 

[37] 

h=S  +  p,  prism. 

[38] 

V  =  Ah,  prism. 

[39] 

F  =  a3,  cube. 

[40] 

h  =  V  +A,  prism. 

[41] 

A  =  F-T-/I,  prism. 

[42] 

S  =  Ch  =Trdh=2irrh,  cylinder. 

[43] 

V  =  Ah  =  wr2h  =  \Trd*h,  cylinder. 

[44] 

h  =  V  -T-  A  =  V  -s-irr2,  cylinder. 

[45] 

A  =  V  -v-  h,  cylinder. 

[46] 

V  =-irR2h  —irr^h  =irh(R2  -r2)  =vh(R  +r)  (R-r),  cylinder. 

[47] 

S  =  %ps,  pyramid  or  cone. 

[48] 

T  =  %ps-{-A,  pyramid  or  cone. 

[49] 

S  =  i(P-fp)Xs,  frustum. 

[50] 

T  =  l(P+p)Xs+B+b,  frustum. 

[51] 

V  =  lAh,  pyramid  or  cone. 

[52] 

V  =  %h(B+b+\/Bb),  frustum  of  pyramid  or  cone. 

[53] 

V  =  %irh(R°+r2+Rr),  frustum  of  cone. 

[54] 

V=&Trh(D2+d2+Dd),  frustum  of  cone. 

[55] 

S  =  4irr2,  sphere. 

[56] 

V  —  \Sr  =  |7ir3  =  gird3,  sphere. 

[57] 

Z  =2irrh,  area  of  zone. 

[58] 

F  =  57r/t(r2i+i"22)+g7r/i3,  volume  of  segment. 

[59] 

A  =2wRX2Trr  =  4TT2Rr,  solid  ring. 

[60] 

F  =  27r#X7rr2=27r2.Rr2,  solid  ring. 

[61] 

V  =  ±h(Bl+4M+B2'),  prismatoid. 

[62] 

sin  (90°  -0)  =cos  6. 

[63] 

cos  (90°  —  0)  =sin  6. 

[64] 

tan  (90°  -0)=  cot  d. 

[65] 

cot  (90°-  e)=  tan  d. 

[66] 

sec  (90°  -0)  =csc  0. 

[67] 

esc  (90°  -0)=  sec  9. 

[68] 

sin  (180°  -6}  =sin  9. 

[69] 

cos  (180°  —  (?)  =  -cos  0. 

[70] 

tan  (  180°  -0)  =  -  tan  0. 

[71] 

cot  (180°  —  0)  =  -cot  0. 

[72] 

sec  (180°  -0)  =  -sec  0. 

[73] 

esc  (180°-<?)=csc  0. 

[74] 

sin  (90°  +  0)=  cos  0. 

[75] 

cos  (90°  +  0)  =  -  sin  0. 

[76] 

tan  (90°  +0)  =  -  cot  0. 

[77] 

cot  (90°  +0)  =  -  tan  0. 

472  PRACTICAL  MATHEMATICS 


[78] 

sec  (90°  +  0)  =  -esc  0. 

[79] 

esc  (90°  -f0)  -sec  0. 

[80] 

sin  (  —  0)  =  —sin  0. 

[81] 

cos  (  —  0)  =  cos  0 

[82] 

tan  (  —  0)  —  —  tan  0. 

[83] 

cot  (—0)  =  —cot  0. 

[84] 

sec  (  —  0)  =sec  0. 

[85] 

esc  (—  0)  =  —esc  0. 

[86] 

1                        1 
sin  0  =  •  or  esc  0  =  -r—  • 
esc  0                   sin  0 

[87] 

1                         1 

cos  0  —        -  or  sec  0  —          • 

sec  0                  cos  0 

[88] 

1                          1 

tan  v  =     .   ..  or  cot  9  —  ,         • 
cot  0                   tan  0 

[89] 

vers  0  =  1—  cos  0. 

[90] 

covers  0  =  1  —sin  0. 

[91] 

sin2  0+cos2  0  =  1. 

[92] 

sec2  0  =  1  +tan2  0. 

[93] 

esc2  0  =  1  +cot2  0. 

FQ4.1 

tan  0-8in<?. 

[y±] 

tan  v  —         • 

COS0 

,      , 

^  0    cos  0 

"•      ' 

sin  0 

[96] 

sin  (A+B)  =sin  A  cos  B+cos  A  sin  B. 

[97] 

sin  (A  —  B)  =sin  A  cos  B—  cos  A  sin  B. 

[98] 

cos  (A+B)  =cos  A  cos  B—  sin  A  sin  B. 

[99] 

cos  (A  —  B)  =cos  A  cos  B+sin  A  sin  B. 

[100] 

/  4  i  z>\       tan  A  +tan  B 

1  1  1  1  1     (  ^.  —f~  ,/J  )  ^s  ~  ^* 

[101] 

tan  (A     B)       tan  A~tan  B 

1+tanAtanB 

[102] 

sin  2A  =2  sin  A  cos  A. 

[103] 

cos  2A=cos2  A  -sin2  A  =1—2  sin2  A  =2  cos2  A-l 

[104] 

tan  o  ,  _    2  tan  A 

tall    £t£\  —    - 

1—  tan2  A 

[105] 

/I  -cos  2A 

»|n          ^      _I_       .4      I 

sin  A  •    :  "\/        9 

'         ^ 

[106] 

/I  +cos  2A 

f*r\a     a    —  <4-    •*    /                                      • 

-  ^ 

[107] 

11  -cos  2A      sin  2  A         1  -cos  2  A 

'  V  1  +cos  2  A  :  "  1  +cos  2  A        sin  2  A 

[108] 

sin  A  cos  B  =  i  sin  (A  +B)  +}  sin  (A  -B). 

[109] 

cos  A  sin  B  =  i  sin  (A  +B)  —  i  sin  (A  —  B). 

[110] 

cos  A  cos  B  =  ^  cos  (A  +B)  +  i  cos  (  A  —  B). 

[111] 

sin  A  sin  B  =  —  §  cos  (A  +B)  +  J  cos  (A  —  B). 

[112] 

sin  A+sin  B=2sin  i(A+B)  cos  J(A-B). 

[113] 

sin  A  —sin  B  =2  cos  j(A  +B)  sin  §(A  —  B). 

1114] 

cos  A  +cos  B  =2  cos  \(A  +B)  cos  $(A  —  B). 

1115] 

cos  A  —cos  B  =  —  2  sin  i(A  +B)  sin  j(A  —  B). 

TABLES  473 


sin  A     sin  B     sin  C 
[117]      a2  =  b2+c2-26ccos4. 
[118]      62=o2+c2-2accos  B. 
[119]      c2=a2+62— 2ab  cos  C. 

7,2_|_,.2_,72 

[120]       cosA=     +0.         ' 
Zbc 

/72_l_r2_7,2 

[1211   ...B-tjiLj-L 

[122]      cos  C  =  " 


2a6 
Formulas  for  solving  Case  III,  oblique  triangles,  by  logarithms: 


[123J      tan  \(B-C)  =j-     tan 

0   |~C 

[124]      tan  $(A-C)  =^-  tan  l(A+C). 
a  i~c 

[125]      tan  hA  -5)=          tan 


Formulas  for  solving  Case  IV,  oblique  triangles,  by  logarithms  : 

a+b+c 
Let  s  = 


[130]      cos  \B  = 
[131]      cos  |C  = 
Let  r 


1R--    /•(«-&). 


,  /(8-q)(«-b)(«-c) 
-\-         — p- 


[132]      tan  *A=-— • 
s— a 

[133]       tan  |/?  = -%- • 
s— o 

[134]      tanJC  =  -^-- 
s— c 

[135]      Area  of  a  triangle  =  5  bcsin  A  =  \ac  sin  B  =  %ab  sin  C 

=  \/s(s  —  a)(s  — 6)  (s  —  c). 
Other  formulas  for  any  angle. 

tan  6  1 

A/1  +tanT0       VHhcoV-  0 


474  PRACTICAL  MATHEMATICS 

1  cote 


\/l+tan»0      Vl+cotf0 

1 


BCC  0  OM   " 


[1371      cos  0~  Vl-sm'0 


U88,      tan  g.  -JEL 

V  1  —sin*  0 


[139]      cot  0 


[140]      sec  0  = 


[141]      csc0--      -     == 

sin  6       V  1— cos2  0 


Vaec*0-l 
[142]      sin  3A  =3  sin  A  -4  sin3  A. 
[143]      cos  ZA  =4  cos3  A  -3  cos  A. 

3  tan  A  —tan3  A 
[144]      tan  34  =   — - — 


TABLE  II 
Useful  Numbers 

1  cu.  ft.  of  water  weighs  62.5  Ib.  (approx.)  =  1000  oz. 

1  gal.  of  water  weighs  8§  Ib.  (approx.). 

1  atmosphere  pressure  =  14.7  Ib.  per  sq.  in.  =2116  Ib.  per  sq.  ft. 

1  atmosphere  pressure  =  760  mm.  of  mercury. 

A  column  of  water  2.3  ft.  high  =a  pressure  of  1  Ib.  per  sq.  in. 

1  gal.  =231  cu.  in.  (by  law  of  Congress). 

1  cu.  ft.  =7J  gal.  (approx.)  or  better  7.48  gal. 

1  cu.  ft.  =  j  bu.  (approx.). 

1  bbl.  =4.211-  cu.  ft.  (approx.). 

1  bu.  =2150.42  cu.  in.  (by  law  of  Congress)  =  1.24446-  cu.  ft. 

1  bu.  =  J  cu.  ft.  (approx.). 

1  perch  =24}  cu.  ft.  but  usually  taken  25  cu.  ft. 

1  in.  =25.4001  mm.  (approx.). 

1  ft.  =-30.4801  cm. 

1  m.  =39.37  in.  (by  law  of  Congress). 

1  lh.  (avoirdupois)  =  7000  grains  (by  law  of  Congress). 


TABLES 

1  Ib.  (troy  or  apothecaries)  =  5760  grains. 

1  gram  =  15.432  grains. 

1  kg.  =2.20462  Ib.  (avoirdupois). 

I  liter  =  1.05668  qt.  (liquid)  =0.90808  qt.  (dry). 

1  qt.  (liquid)  =946.358  cc.  =0.946358  liter,  or  cu.  dm. 

1  qt.  (dry)  =1101.228  cc.  =1.101228  liters,  or  cu.  dm. 

TT=  3. 14159265358979  +  =  3.1416  =  fB  =  3f  (all  approx.). 

1  radian  =  57°  17'  44.8"  =  57.2957795°+. 

1°  =  0.01745329+  radian. 

Base  of  Napierian  logarithms  =e  =2.718281828  •     •     •    . 

log  10  e=  0.43429448   •    •    • 

log  JO  =2. 30258509   •     •     •    . 

1  horse-power  second  =  550  foot-pounds. 

1  horse-power  minute  =  33,000  foot-pounds. 

V2  =  1.4142136.  V3  =  1.7320508. 

\/5  =  2.2360680.  \/6  =  2. 4494897. 

^2  =  1.2599210.  -^3  =  1.4422496. 


475 


TABLE  III 
Decimal  and  Fractional  Equivalents  of  Parts  of  an  Inch 


8ths 

32nds 

64ths 

1  =  .  125 

1  =  .  03125 

1  =  .  015625 

33  =.515625 

2  =.250 

3  =  .09375 

3  =.046875 

35  =.546875 

3  =  .  375 

5  =  .15625 

5  =.078125 

37  =.578125 

4  =  .  500 

7  =.21875 

7  =  .  109375 

39  =  .  609375 

5  =  .  625 

9  =.28125 

9  =  .  140625 

41  =  .  640625 

6  =  .  750 

11  =  .34375 

11  =  .171875 

43  =  .  671875 

7  =  .  875 

13  =.40625 

13  =.203125 

45  =.703  125 

16ths 

15  =  .  46875 

15  =  .  234375 

47  =  .  734375 

1  =  .  0625 

17  =.53125 

17  =  .  265625 

49  =  .  765625 

3  =.1875 

19  =  .  59375 

19  =  .  296875 

51  =  .  796875 

5  =.3125 

21  =  .  65625 

21  =  .328125 

53  =  .  828125 

7  =  .4375 

23  =.71875 

23  =  .  359375 

55  =  .  859375 

9  =  .  5625 

25  =.78125 

25  =  .  390625 

57  =  .  890625 

11  =  .6875 

27  =  .  84375 

27  =.421875 

59  =  .  921875 

13  =  .8125 

29  =.90625 

29  =  .453125 

61  =  .953125 

15  =  .9375 

31  =  .96875 

31  =  .484375 

63  =  .984375 

476 


I'll  A  C'TICAL  MA  THEM  A  TICS 


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O'*  oo  e«  «o  o  ^«  oo  ci  i  o  **  oo  M  «o  o^t"  oo  cNloo  •*  OO~CN       e 

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TABLES 


477 


TABLE  V 
U.  S.  Standard  and  Sharp  V-threads 


Diameter  of  screw 

Threads 
per  inch 

Depth 
U.  S.  S. 

Depth 
sharp  V 

Root  dia. 
U.  S.  S. 

Root  dia. 
sharp  V 

i 

20 

.  03247 

.04330 

.1850 

.1634 

tV 

18 

.03608 

.04811 

.2403 

.2163 

t 

16 

.  04059 

.05412 

.2936 

.2668 

A 

14 

.04639 

.06178 

.3447 

.3139 

* 

13 

.  04996 

.  06661 

.4001 

.3668 

A 

12 

.05412 

.07216 

.4542 

.4182 

f 

11 

.  05905 

.07873 

.5069 

.4675 

i 

10 

.06495 

.08660 

.6201 

.5768 

1 

9 

.07216 

.09622 

.7307 

.6826 

i 

8 

.08119 

.  10825 

.8376 

.7835 

li 

7 

.09277 

.12371 

.9394 

.8776 

H 

7 

.09277 

.12371 

1.0644 

1.0026 

H 

6 

.  10825 

.  14433 

1.1585 

1.0863 

U 

6 

.  10825 

.  14433 

1  .  2835 

1.2113 

l! 

5£ 

.11809 

.  15745 

1  .  3888 

1.3101 

if 

5 

.  12990 

.  17325 

.1  .  4902 

1  .  4035 

U 

5 

.  12990 

.  17325 

1.6152 

1  .  5285 

2 

4J 

.  14433 

.  19244 

1.7113 

1.6151 

2J 

4£ 

.  14433 

.  19244 

1.9613 

1.8651 

2J 

4 

.  16238 

.21650 

2.1752 

2.0670 

2f 

4 

.  16238 

.21650 

2.4252 

2.3170 

3 

3J 

.  18557 

.  24742 

2.6288 

2.5052 

3} 

3J 

.  18557 

.  24742 

2.8788 

2.7552 

3^ 

31 

.  19985 

.  26647 

3.1003 

2.9671 

3| 

3 

.21666 

.  28866 

3.3167 

3.1727 

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.  21666 

.28866 

3.5667 

3.4227 

4J 

21 

.2259 

.3012 

3  .  7982 

3.6476 

4* 

21 

.2362 

.3149 

4.0276 

3.8712 

4f 

2f 

.2474 

.3299 

4.2551 

4.0901 

5 

2-1 

.2598 

.3465 

4.4804 

4.3070 

51 

21 

.2598 

.3465 

4.7304 

4.5500 

5*                                2| 

.2735 

.3647 

4.9530 

4.7707 

51                               2f 

.  2735 

.3647 

5.2030 

5.0207 

6 

2J 

.2887 

.3849 

5.4226 

5.2302 

478 


PRACTICAL  MATHEMATICS 


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TABLES 


479 


TABLE  VII 

Standard  Gages  for  Wire  and  Sheet  Metals 

Diameter  or  thickness  given  in  decimals  of  an  inch 


Number 
of  gage 

Birmingham 
wire  gage 

American, 
Brown  and 
Sharp 
(B.  and  S.) 

United  States 
standard  plate 
iron  steel 

British 
Imperial 

American 
Steel  and 
Wire  Co. 

3000000 

.5 

.5 

000000 

.46875 

.464 

00000 

.4375 

.432 

0000 

.454 

.46 

.40625 

.400 

.3938 

000 

.425 

.409642 

.375 

.372 

.3625 

00 

.380 

.364796 

.34375 

.348 

.3310 

0 

.340 

.324861 

.3125 

.324 

.3065 

1 

.300 

.  289297 

.28125 

.300 

.2830 

2 

.284 

.257627 

.265625 

.276 

.2625 

3 

.259 

.  229423 

.25 

.252 

.2437 

4 

.238 

.204307 

.234375 

.232 

.2253 

5 

.220 

.  181940 

.21875 

.212 

.2070 

•   6 

.203 

.  162023 

.203125 

.192 

.1920 

7 

.180 

.  144285 

.1875 

.176 

.1770 

8 

.165 

.  128490 

.171875 

.160 

.1620 

9 

.148 

.114423 

.  15625 

.144 

.1483 

10 

.134 

.101897 

.  140625 

.128 

.1350 

11 

.120 

.  090742 

.125 

.116 

.1205 

12 

.109 

.  080808 

.  109375 

.104 

.1055 

13 

.095 

.071962 

.09375 

.092 

.0915 

14 

.083 

.  064084 

.078125 

.080 

.0800 

15 

.072 

.  057068 

.0703125 

.072 

.0720 

16 

.065 

.050821 

.0625 

.064 

.0625 

17 

.058 

.  045257 

.05625 

.056 

.0540 

18 

.049 

.040303 

.05 

.048 

.0475 

19 

.042 

.035890 

.04375 

.040 

.0410 

20 

.035 

.031961 

.0375 

.036 

.0348 

21 

.032 

.028462 

.034375 

.032 

.03175 

22 

.028 

.  025346 

.03125 

.028 

.0286 

23 

.025 

.022572 

.028125 

.024 

.0258 

24 

.022 

.020101 

.025 

.022 

.0230 

25 

.020 

.017900 

.021875 

.020 

.0204 

26 

.018 

.015941 

.01875 

.018 

.0181 

27 

.016 

.014195 

.0171875 

.0164 

.0173 

28 

.014 

.012641 

.015625 

.0148 

.0162 

29 

.013 

.011257 

.0140625 

.0136 

.0150 

30 

.012 

.010025 

.0125 

.0124 

.0140 

31 

.010 

.008928 

.0109375 

.0116 

.0132 

32 

.009 

.007950 

.01015625 

.0108 

'  .0128 

33 

.008 

.007080 

.009375 

.0100 

.0118 

34 

.007 

.  006305 

.00859375 

.0092 

.0104 

35 

.005 

.005615 

.0078125 

.0084 

.0095 

36 

.004 

.005000 

.00703125 

.0076 

.0090 

37 

.004453 

.006640625 

.0068 

38 

.003965 

.00625 

.0060 

39 

.003531 

.0052 

40 

.003144 

.0048 

480  PRACTICAL  MATHEMATICS 

TABLE  VIII 
Specific  Gravities  and  Weights  of  Substances 


.Vamc  of  substance 

Pounds  per 

en.  in. 

Pounds  per 
cu.  ft. 

Specific 
gravity 

Air  

0.0795 

Aluminum  

162 

2   6 

Anthracite  coal,  broken.     .    .  . 

52  to  60 

Antimony  

418 

6  7 

Asphaltum  

87.3 

1  4 

Beech  wood  

46 

.73 

Birch  wood  .  .                    

41 

65 

Brass,  cast  (copper  and  zinc)  .... 

506 

8  1 

Brass,  rolled  

525 

8  4 

Brick,  common  

125 

Brick,  pressed                      .    ... 

150 

Chalk  

156 

2  5 

Coal,  bituminous,  broken  

47  to  56 

Coke,  loose 

23  to  32 

Corundum           .  .            

3  9 

Copper,  cast  

542 

8  .  6  to  8  8 

Copper,  rolled  

.319. 

555 

8  .  8  to  9 

Cork 

15 

24 

Ebony  wood 

76 

1  23 

Elm  wood  

35 

.56 

Flint 

162 

2  6 

Glass                                       

186 

2  5  to  3  45 

Gold        

.695 

19  3 

Granite  

170 

2.56  to2.88 

Hickory  wood                           .  . 

53 

85 

Ice  

57.5 

.92 

Iron,  cast. 

.26 

450 

6.7  to  7.4 

Iron,  wrought.    .  .            

.28 

480 

7.69 

Lead 

.412 

712 

11  42 

Marble     •                                  .... 

168.7 

2.7 

Maple  wood                                     • 

49 

79 

Mercury.  .         

.49 

13.6 

Nickel  

.318 

8.8 

Oak  wood,  red                           .... 

46 

.73  to  .75 

Pine  wood    white                           . 

28 

.45 

Pine  wood   yellow                      .    . 

38 

.61 

Platinum  .  .                           .... 

21.5 

Quartz  

165 

2.65 

Silver                                      

.379 

655 

10.5 

TABLES 


481 


Specific  Gravities  and  Weights  of  Substances — Continued 


Name  of  substance 

Pounds  per 
cu.  in. 

Pounds  per 
cu.  ft. 

Specific 
gravity 

Steel 

.29 

490 
459 
438 
62.417 
62  .  355 
59.7 

7.85 
7  .  2  to  7  .  5 
6  .  8  to  7  .  2 

1 

Tin 

Zinc                  .  . 

Water,  distilled 
Water,  distilled 
Water,  distilled 

at  32°  F.    . 

at  62°  F 

at  212°  F 

Specific  gravities  referred  to  air : 

Air 1 

Oxygen 1.11 

Hydrogen 0 . 07 

Chlorine  gas 2 . 44 

TABLE  IX 
Strength  of  Materials 


Ultimate  tensile 
strength 

Ultimate  com- 
pressive  strength 

Coefficient    of 
linear  expansion 

Pounds  per 
square  inch 

Pounds  per 
square  inch 

For  1° 
Fahrenheit 

Hard  steel 

100,000 

120,000 

0  0000065 

Structural  steel  

60,000 

60,000 

0  .  0000065 

Wrought  iron 

50,000 

50,000 

0  0000067 

Cast  iron 

20,000 

90,000 

0  0000062 

Copper  ...          .        .... 

30,000 

0  .  0000089 

Timber,  with  grain  
Concrete  

8,000  to 
25,000 
300 

4,000  to 
12,000 
3,000 

0.0000028 
0  0000055 

Granite.  ...        .  .        .    . 

11,000 

0  0000050 

Brick  

3,000 

0  0000050 

In  the  column  of  ultimate  tensile  strengths  are  given  the  pulls  neces- 
sary to  break  a  rod  of  one  square  inch  cross  section  of  the  given  material. 

In  the  column  of  ultimate  compressive  strengths  are  given  the  weights 
necessary  to  cause  a  support  of  one  square  inch  cross  section  to  give  way 
under  the  pressure. 

In  the  column  of  coefficient  of  linear  expansion  are  given  the  fractional 
parts  of  their  length,  bars  of  the  different  materials  will  increase  when  the 
temperature  rises  one  degree  Fahrenheit. 

31 


482 


PRACTICAL  MATHEMATICS 

TABLE  X.— COMMON  LOGARITHMS 


t)    , 

. 

. 

1  1 

10 

0000 

0043 

0086 

0128 

0170 

0212 

0253 

0294 

0334 

0374 

11 

0414 

0453 

0492 

0531 

0569 

0607 

0645 

0682 

0719 

0755 

1* 

0792 

0828 

0864 

0899 

0934 

ii'.tii'.i 

1004 

1038 

1072 

1106 

13 

1139 

1173 

1206 

1239 

1271 

1303 

1335 

1367 

1399 

1430 

14 

1461 

1492 

1523 

1553 

.1584 

1614 

1644 

1673 

1703 

1732 

16 

1761 

1790 

1818 

1847 

1875 

1903 

1931 

1959 

1987 

2014 

16 

2041 

2068 

2095 

2122 

2148 

2175 

2201 

2227 

2253 

2279 

17 

2304 

2330 

2355 

2380 

2405 

2430 

2455 

2480 

2504 

2529 

18 

2553 

2577 

2601 

2625 

2648 

2672 

2695 

2718 

2742 

ITU 

19 

2788 

2810 

2833 

2856 

2878 

2900 

2923 

2945 

2967 

2989 

10 

3010 

3032 

3054 

3075 

3096 

3118 

3139 

3160 

3181 

3201 

21 

3222 

3243 

3263 

3284 

3304 

3324 

3345 

3365 

3385 

3404 

22 

3424 

3444 

3464 

3483 

3502 

3522 

3541 

3560 

3579 

UM 

23 

3617 

3636 

3655 

3674 

3692 

3711 

3729 

3747 

3766 

3784 

24 

3802 

3820 

3838 

3856 

3874 

3892 

3909 

3927 

3945 

3962 

25 

3979 

3997 

4014 

4031 

4048 

4065 

4082 

4099 

4116 

4133 

26 

4150 

4166 

4183 

4200 

4216 

4232 

4249 

4265 

4281 

4298 

27 

4314 

4330 

4346 

4362 

4378 

4393 

4409 

4425 

4440 

4456 

28 

4472 

4487 

4502 

4518 

4533 

4548 

4564 

4579 

4594 

4609 

29 

4624 

4639 

4654 

4669 

4683 

4698 

4713 

4728 

4742 

4757 

80 

4771 

4786 

4800 

4814 

4829 

4843 

4857 

4871 

4886 

4900 

31 

4914 

4928 

4942 

4955 

4969 

4983 

4997 

5011 

5024 

5038 

32 

5051 

5065 

5079 

5092 

5105 

5119 

5132 

5145 

5159 

5172 

33 

6185 

5198 

5211 

5224 

5237 

5250 

5263 

5276 

5380 

5302 

34 

5315 

5328 

5340 

5353 

5366 

5378 

5391 

5403 

5416 

5428 

36 

5441 

5453 

5465 

5478 

5490 

5502 

5514 

5527 

5539 

5551 

36 

5563 

5575 

5587 

5599 

5611 

5623 

5635 

5647 

5658 

5670 

37 

5682 

5694 

5705 

5717 

5729 

5740 

5752 

5763 

5775 

5786 

38 

5798 

5809 

5821 

5832 

5843 

5855 

5866 

5877 

5888 

5899 

39 

5911 

5922 

5933 

5944 

5955 

5966 

5977 

5988 

5999 

6010 

40 

6021 

6031 

6O42 

6053 

6064 

6075 

6085 

6096 

6107 

6117 

41 

6128 

6138 

6149 

6160 

6170 

6180 

6191 

6201 

6212 

6222 

42 

6232 

6243 

6253 

6263 

6274 

fC'Sl 

6294 

6304 

6314 

6325 

43 

6335 

6345 

6355 

6365 

6375 

6385 

6395 

6405 

6415 

6425 

44 

6435 

6444 

i.l.  ->1 

6464 

6474 

6484 

6493 

6503 

6513 

6522 

46 

6532 

6542 

6551 

6561 

6571 

6580 

6590 

6599 

6609 

6618 

46 

6628 

6637 

6646 

6656 

6665 

6675 

6684 

6693 

6702 

6712 

47 

6721 

6730 

6739 

6749 

6758 

6767 

6776 

6785 

6794 

BOH 

48 

6812 

6821 

6830 

6839 

6848 

6857 

6866 

6875 

6884 

6893 

49 

6902 

6911 

6920 

6928 

6937 

6946 

6955 

6964 

6972 

6981 

60 

6990 

6998 

7007 

7016 

7024 

7033 

7042 

7050 

7059 

7067 

61 

7070 

7084 

7093 

7101 

7110 

7118 

7126 

7135 

7143 

7152 

62 

7160 

7168 

7177 

7185 

7193 

7202 

7210 

7218 

7226 

7235 

63 

7243 

7251 

7259 

7267 

7275 

7284 

7292 

7300 

7308 

7316 

64 

7324 

7332 

7340 

7348 

7350 

7364 

7372 

7380 

7388 

7396 

N. 

0 

1 

2 

3 

4 

9 

6 

7 

8 

9 

TABLES 

TABLE  X.— COMMON  LOGARITHMS—  Continued 


483 


N. 

0 

1 

2 

3 

4 

0 

6 

7 

8 

9 

55 

7404 

7412 

7419 

7427 

7435 

7443 

7451 

7459 

7466 

7474 

66 

7482 

7490 

7497 

7505 

7513 

7520 

7528 

7536 

7543 

7551 

57 

7559 

7566 

7574 

7582 

7589 

7597 

7604 

7612 

7619 

7627 

58 

7634 

7642 

7649 

7657 

7664 

7672 

7679 

7686 

7694 

7701 

59 

7709 

7716 

7723 

7731 

7738 

7745 

7752 

7760 

7767 

7774 

60 

7782 

7789 

7796 

7803 

7810 

7818 

7825 

7832 

7839 

7846 

61 

7853 

7860 

7868 

7875 

7882 

7889 

7896 

7903 

7910 

7917 

62 

7924 

7931 

7938 

7945 

7952 

7959 

7966 

7973 

7980 

7987 

63 

7993 

8000 

8007 

8014 

8021 

8028 

8035 

8041 

8048 

8055 

64 

8062 

8069 

8075 

8082 

8089 

8096 

8102 

8109 

8116 

8122 

65 

8129 

8136 

8142 

8149 

8156 

8162 

8169 

8176 

8182 

8189 

66 

8195 

8202 

8209 

8215 

8222 

8228 

8235 

8241 

8248 

8254 

67 

8261 

8267 

8274 

8280 

8287 

8293 

8299 

8306 

8312 

8319 

68 

8325 

8331 

8338 

8344 

8351 

8357 

8363 

8370 

8376 

8382 

69 

8388 

8395 

8401 

8407 

8414 

8420 

8426 

8432 

8439 

8445 

70 

8451 

8457 

8463 

8470 

8476 

8482 

8488 

8494 

8500 

8506 

71 

8513 

8519 

8525 

8531 

8537 

8543 

8549 

8555 

8561 

8567 

72 

8573 

8579 

8585 

8591 

8597 

8603 

8609 

8615 

8621 

8627 

73 

8633 

8639 

8645 

8651 

8657 

8663 

8669 

8675 

8681 

8686 

74 

8692 

8698 

8704 

8710 

8716 

8722 

8727 

8733 

8739 

8745 

75 

8751 

8756 

8762 

8768 

8774 

8779 

8785 

8791 

8797 

8802 

76 

8808 

8814 

8820 

8825 

8831 

8837 

8842 

8848 

8854 

8859 

77 

8865 

8871 

8876 

8882 

8887 

8893 

8899 

8904 

8910 

8915 

78 

8921 

8927 

8932 

8938 

8943 

8949 

8954 

8960 

8965 

8971 

79 

8976 

8982 

8987 

8993 

8998 

9004 

9009 

9015 

9020 

9025 

80 

9031 

9036 

9042 

9047 

9053 

9058 

9063 

9069 

9074 

9079 

81 

9085 

9090 

9096 

9101 

9106 

9112 

9117 

9122 

9128 

9133 

82 

9138 

9143 

9149 

9154 

9159 

9165 

9170 

9175 

9180 

9186 

83 

9191 

9196 

9201 

9206 

9212 

9217 

9222 

9227 

9232 

9238 

84 

9243 

9248 

9253 

9258 

9263 

9269 

9274 

9279 

9284 

9289 

85 

9294 

9299 

9304 

9309 

9315 

9320 

9325 

9330 

9335 

9340 

86 

9345 

9350 

9355 

9360 

9365 

9370 

9375 

9380 

9385 

9390 

87 

9395 

9400 

9405 

9410 

9415 

9420 

9425 

9430 

9435 

9440 

88 

9445 

9450 

9455 

9460 

9465 

9469 

9474 

9479 

9484 

9489 

89 

9494 

9499 

9504 

9509 

9513 

9518 

9523 

9528 

9533 

9538 

90 

9542 

9547 

9552 

9557 

9562 

9566 

9571 

9576 

9581 

9586 

91 

9590 

9595 

9600 

9605 

9609 

9614 

9619 

9624 

9628 

9633 

92 

9638 

9643 

9647 

9652 

9657 

9661 

9666 

9671 

9675 

9680 

93 

9685 

9689 

9694 

9699 

9703 

9708 

9713 

9717 

9722 

9727 

94 

9731 

9736 

9741 

9745 

9750 

9754 

9759 

9763 

9768 

9773 

95 

9777 

9782 

9786 

9791 

9795 

9800 

9805 

9809 

9814 

9818 

96 

9823 

9827 

9832 

9836 

9841 

9845 

9850 

9854 

9859 

9863 

97 

9868 

9872 

9877 

9881 

9886 

9890 

9894 

9899 

9903 

9908 

98 

9912 

9917 

9921 

9926 

9930 

9934 

9939 

9943 

9948 

9952 

99 

9956 

9961 

9965 

9969 

9974 

9978 

9983 

9987 

9991 

9996 

N. 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

PRACTICAL  MATHEMATICS 


TABLE  XI.— TRIGONOMETRIC  FUNCTIONS 


Angle* 

Sine* 

Cosines 

Tangents 

Cotangents 

AngW 

Nat.      Log. 

Nat.      Log. 

Nat.      Log. 

Nat.         Log. 

OeOO' 

.0000      a, 

1  .  0000  0  OOOO 

.0000      OP 

CD                                 00 

90°  00' 

10 

.0029  7.4637 

1.0000       0000 

.0029  7.4637 

343.77       2.5363 

50 

20 

.0058       7048 

1.0000       OOOO 

.0058       7648 

171.89           2352 

40 

30 

.0087       9408 

1.0000       OOOO 

.0087       9409 

114.59           0591 

30 

40 

.0116  8.0658 

.9999       OOOO 

.0116  8.0658 

85.940     1.9842 

20 

50 

.0145       1627 

.9900       OOOO 

.0145       1627 

68  .  750         8373 

10 

1°00' 

.0175  8.2419 

.90M  9.9999 

.0175  8.2419 

57.290     1.7581 

89°  00' 

10 

.0204       3088 

.  9998       9999 

.0204       3089 

49.104         6911 

50 

20 

.0233        3668 

.9997       9999 

.0233       3669 

42.964          6331 

40 

30 

.0262       4179 

.9997       9999 

.0262       4181 

38.188         5819 

30 

40 

.0291       4637 

.  9996       9998 

.0291        4638 

34  .  368         5362 

20 

50 

.  0320       5050 

.9995       9998 

.0320       5053 

31.242         4947 

10 

2°  00' 

.0349  8.5428 

.9994  9.9997 

.0349  8.5431 

28.636     1.4569 

88°  00' 

10 

.0378       5770 

.9993       9997 

.0378       5779 

26.432         4221 

50 

20 

.0407     •  6097 

.9992       9996 

.0407       6101 

24  .  542        3800 

40 

30 

.0430       6397 

.9990       9996 

.0437        6401 

22.904          3599 

30 

40 

.0465       6077 

.9989        9995 

.0466       6682 

21.470          3318 

20 

60 

.0494        6940 

.9988       9995 

.0495       6945 

20.206          3055 

10 

3°  00' 

.0523  8.7188 

.9986  9.9994 

.0524  8.7194 

19.081      1.2806 

87°  00' 

10 

.0552       7423 

.9985       9993 

.0553       7429 

18.075         2571 

50 

20 

.0581        7645 

.9983       9993 

.0582        7652 

17.169         2348 

40 

30 

.0610       7857 

.9981        9992 

.0612       7865 

16.350         2135 

30 

40 

.0640       8059 

.9980       9991 

.0641        8067 

15.605          1933 

20 

60 

.0669       8251 

.9978       9990 

.0670       8261 

14.924          1739 

10 

4°00/ 

.0098  8.8436 

.9976  9.9989 

.0699  8.8440 

14.301      1.1554 

86°  00' 

10 

.0727        8613 

.9974        9989 

.0729        8024 

13.727          1376 

50 

20 

.0756        8783 

.9971        9988 

.0758       8795   i    13.197          1205  i           40 

30 

.0785       8946 

.9909       9987 

.0787        8960       12.700          1040  1           30 

40 

.0814        9104 

.  9967        9980 

.0810       9118 

12.251          0882 

20 

50 

.0843       9256 

.9964        9985 

.0840       9272 

11.820         0728 

10 

5°  00' 

.0872  8.9403 

.9902  9.9983 

.0875  8.9420 

11.430     1.0680 

85°  00' 

10 

.0901        9545 

.9959       9982 

.0904        9503 

11.059         0437 

50 

20 

.0929       9682 

.9957       9981 

.0934       9701 

10.712         0299 

40 

30 

.0958       9816 

.9954        9980 

.0963       9836 

10.385         0164 

30 

40 

.0987       9945 

.9951       9979 

.0992       9966 

10.078         0034 

20 

60 

.1016  9.0070 

.9948     -9977 

.1022  9.0093 

9.7882  0.9907 

10 

6°  00' 

.1045  9.0192 

.9945  9.9976 

.1051   9.0216 

9.5144  0.9784 

84°  00' 

10 

.1074       0311 

.9942       9975 

.  1080       0336 

9.2553       9664 

50 

20 

.1103       0426 

.9939       9973 

.1110       0453 

9.0098       9547 

40 

30 

.1132       0539 

.9936       9972 

.1139       0567 

8.7769       9433 

30 

40 

.1161        0648 

.9932       9971 

.1169       0678 

8.5555       9322 

20 

50 

.  1  190       0755 

.9929       9969 

.1198       0786 

8.3450       9214 

10 

7°  00' 

.1219  9.0859 

.9925  9.9968 

.1228  9.0891 

8.1443  0.9109 

83°  OO7 

10 

.1248       0961 

.9922       9966 

.1257       0995 

7.9530     -9006 

50 

20 

.1276        1060 

.9918       9964 

.1287       1096 

7.7704       8904 

40 

30 

.1305        1157 

.9914        9963 

.1317        1194 

7.5958       8806 

30 

40 

.1334        1252 

.9911        9961 

.1346        1291 

7.4287       8709 

20 

50 

.  1363        1345 

.9907       9959 

.1376        1385 

7.2687       8615 

10 

8°00/ 

.1392  9.1436 

.9903  9.9958 

.1405  9.1478 

7.1154  0.8522 

82°  00' 

10 

.1421        1525 

.9899       9956 

.  1435        1509 

0.9082       8431 

50 

20 

.1449        1612 

.9894        9954 

.1465        1658 

0.8269       8342 

40 

30 

.  1478       1697 

.  9890       9952 

.1495        1745 

6.0912       8255 

30 

40 

.1507       1781 

.9886       9950 

.1524        1831 

6.5600       8169 

20 

50         .1536        1863 

.9881        9948 

.1554        1915 

6.4348       8085 

10 

9°  00'        .1504  9.1943   i    .9877  9.9940 

.1584  9    1997 

6.3138  0.8003 

81°  00' 

Nat.      Log.          Nat.      Log. 

Nat.      Log. 

Nat.         Log. 

Angle* 

Cosines 

Sines 

Cotangents 

Tangents 

Angles 

TABLES 


485 


TABLE  XL— TRIGONOMETRIC  FUNCTIONS—  Continued 


Angles 

Sines 

Cosines 

Tangents 

Cotangents 

Angles 

Nat.   Log. 

Nat.   Log. 

Nat.   Log. 

Nat.    Log. 

9°  00' 

.1564  9.1943 

.9877  9.9946 

.1584  9.1997 

6.3138  0.8003 

81°  00' 

10 

.1593   2022 

.9872   9944 

.  1614   2078 

6.1970   7922 

50 

20 

.  1622   2100 

.9868   9942 

.1644   2158 

6.0844   7842 

40 

30 

.1650   2176 

.  9863   9940 

.  1673   2236 

5.9758   7764 

30 

40 

.1679   2251 

.9858   9938 

.  1703   2313 

5.8708   7687 

20 

50 

.  1708   2324 

.9853   9936 

.  1733   2389 

5.7694  .  7611 

10 

10°  00' 

.1736  9.2397 

.9848  9.9934 

.1763  9.2463 

5.6713  0.7537 

80°  00' 

10 

,  .1765   2468 

.9843   9931 

.  1793   2536 

5.5764   7464 

50 

20 

.  1794   2538 

.  9838   9929 

.  1823   2609 

5.4845   7391 

40 

30 

.  1822   2606 

.9833   9927 

.1853   2680 

5.3955   7320 

30 

40 

.  1851   2674 

.9827   9924 

.  1883   2750 

5.3093   7250 

20 

50 

.  1880   2740 

.9822   9922 

.1914   2819 

5.2257   7181 

10 

11°  00' 

.1908  9.2806 

.9816  9.9919 

.1944  9.2887 

5.1446  0.7113 

79°  00' 

10 

.  1937   2870 

.9811   9917 

.  1974   2953 

5.0658   7047 

50 

20 

.  1965   2934 

.9805   9914 

.  2004   3020 

4.9894   6980 

40 

30 

.  1994   2997 

.9799   9912 

.2035   3085 

4.9152   6915 

30 

40 

.  2022   3058 

.  9793   9909 

.2065   3149 

4.8430   6851 

20 

50 

.2051   3119 

.  9787   9907 

.2095   3212 

4.7729   6788 

10 

12°  00' 

.2079  9.3179 

.9781  9.9904 

.2126  9.3275 

4.7046  0.6725 

78°  00' 

10 

.2108   3238 

.9775   9901 

.2156   3336 

4  .  6382   6664 

50 

20 

.2136   3296 

.9769   9899 

.2186   3397 

4  .  5736   6603 

40 

30 

.2164   3353 

.9763   9896 

.2217   3458 

4.5107   6542 

30 

40 

.2193   3410 

.9757   9893 

.2247   3517 

4.4494   6483 

20 

50 

.2221   3466 

.  9750   9890 

.2278   3576 

4.3897   6424 

10 

13°  00' 

.2250  9.3521 

.9744  9.9887 

.2309  9.3634 

4.3315  0.6366 

77°  00' 

10 

.2278   3575 

.9737   9884 

.2339   3691 

4  .  2747   6309 

50 

20 

.  2306   3629 

.9730   9881 

.2370   3748 

4.2193   6252 

40 

30 

.2334   3682 

.9724   9878 

.  2401   3804 

4.1653   6196 

30 

40 

.2363   3734 

.9717   9875 

.  2432   3859 

4.1126   6141 

20 

50 

.2391   3786 

.9710   9872 

.2462   3914 

4.0611   6086 

10 

14°  00' 

.2419  9.3837 

.9703  9.9869 

.2493  9.3968 

4.0108  0.6032 

76°  00' 

10 

.2447   3887 

.9696   9866 

.2524   4021 

3.9617   5979 

50 

20 

.  2476   3937 

.9689   9863 

.  2555   4074 

3.9136   5926 

40 

30 

.  2504   3986 

.9681   9859 

.2586   4127 

3  .  8667   5873 

30 

40 

.  2532   4035 

.  9674   9856 

.2617   4178 

3.8208   5822 

20 

50 

.  2560   4083 

.9667   9853 

.2648   4230 

3.7760   5770 

10 

15°  00' 

.2588  9.4130 

.9659  9.9849 

.2679  9.4281 

3.7321  0.5719 

75°  00' 

10 

.2616   4177 

.9652   9846 

.2711   4331 

3.6891   5669 

50 

20 

.  2644   4223 

.  9644   9843 

.2742   4381 

3.6470   5619 

40 

30 

.  2672   4269 

.9636   9839 

.  2773   4430 

3  .  6059   5570 

30 

40 

.2700   4314 

.9628   9836 

.  2805   4479 

3.5656   5521 

20 

50 

.2728   4359 

.9621   9832 

.  2836   4527 

3.5261   5473 

10 

16°  00' 

.2756  9.4403 

.9613  9.9828 

.2867  9.4575 

3.4874  0.5425 

74°  00' 

10 

.2784   4447 

.9605   9825 

.  2899   4622 

3.4495   5378 

50 

20 

.2812   4491 

.9596   9821 

.2931   4669 

3.4124   5331 

40 

30 

.  2840   4533 

.9588   9817 

.2962   4716 

3  .  3759   5284 

30 

40 

.2868   4576 

.9580   9814 

.2994   4762 

3.3402   5238 

20 

50 

.2896   4618 

.9572   9810 

.3026   4808 

3.3052   5192 

10 

17°  00' 

.2924  9.4659 

.9563  9.9806 

.3057  9.4853 

3.2709  0.5147 

73°  00' 

10 

.  2952   4700 

.9555   9802 

.  3089   4898 

3.2371   5102 

50 

20 

.2979   4741 

.9546   9798 

.3121   4943 

3.2041   5057 

40 

30 

.3007   4781 

.9537   9794 

.3153   4987 

3.1716   5013 

30 

40 

.3035   4821 

.9528   9790 

.3185   5031 

3.1397   4969 

20 

50 

3062   4861 

.9520   9786 

.3217   5075 

3.1084   4925 

10 

18°  00' 

.3090  9.4900 

.9511  9.9782 

.3249  9.5118 

3.0777  0.4882 

72°  00' 

Nat.   Log. 

Nat.   Log. 

Nat.   Log. 

Nat.   Log. 

Angles 

Cosines 

Sines 

Cotangents 

Tangents 

Angles 

486 


PR  A  CTICAL  MA  THEM  A  TICS 


TABLE  XI.— TRIGONOMETRIC  FUNCTIONS— Continued 


Angle* 

Sine* 

Cosines 

Tangents 

Cotangents 

Angles 

Nat.      Log. 

Nat.      Log. 

Nat.      Log. 

Nat.       Log. 

18°  00' 

.3090  9.4900 

.9511  0.9782 

.3249  9.5118 

3.0777  0.4882 

72°  00' 

10 

.3118       4939 

.9502       9778 

.3281       6161 

3.0475       4839 

60 

20 

.3145       4977 

.9492       9774 

.3314       6203 

3.0178       4797 

40 

30 

.3173       6015 

.9483       9770 

.3346       5245 

S.M87       4755 

30 

40 

.3201       5052 

.9474       9765 

.3378       5287 

2.9600       4713 

20 

60 

.3228       5090 

.9465       9761 

.3411       5329 

2.9319       4671 

10 

19°  00' 

.3256  9.5126 

.9455  9.9757 

.3443  9.5370 

2.9042  0.4630 

71°  00' 

10 

.3283       5163 

.9446       9752 

.3476       5411 

2.8770       4589 

50 

20 

.3311       5199 

.9436       9748 

.3508       5451 

2.8502       4549 

40 

30 

.3338       5235 

.9426       9743 

.3541       5491 

2.8239       4509 

30 

40 

.3365       5270 

.9417       9739 

.3574       5531 

2.7980       4469 

20 

60 

.3393       5306 

.9407       9734 

.3607       5571 

2  .  7725       4429 

10 

20°  00' 

.3420  9.5341 

.9397  9.9730 

.3640  9.5611 

2.7475  0.4389 

70°  0<y 

10 

.  3448       6375 

.9387       9725 

.3673       5650 

2.7228       4350 

60 

20 

.3475       5409 

.9377       9721 

.3706       5689 

2.6985       4311 

40 

30 

.3502        5443 

.9367       9716 

.3739       5727 

2  .  6746       4273 

30 

40 

.3529       5477 

.9356       9711 

.  3772        5766 

2.6511        4234 

20 

60 

.3557       5510 

.9346       9706 

.  3805       5804 

2.6279       4196 

10 

21°  OCX 

.3584  9.5543 

.9336  9.9702 

.3839  9.5842 

2.6051  0.4158 

69°  00' 

10 

.3611        5576 

.9325       9697 

.  3872       5879 

2.0826       4121 

60 

20 

.3638       5609 

.9315       9692 

.3906       5917 

2  .  5605       4083 

40 

30 

.3665       5641 

.9304       9687 

.  3939       5954 

2.5386       4046 

30 

40 

.3692       5673 

.9293       9682 

.3973       5991 

2.5172       4009 

20 

50 

.3719       5704 

.9283       9677 

.  4006       6028 

2.4960       3972 

10 

22°  00' 

.3746  9.5736 

.9272  9.9672 

.4040  9.6064 

2.4751  0.3936 

68°  00' 

10 

.3773       5767 

.9261        9667 

.4074       6100 

2.4545       3900 

60 

20 

.  3800       5798 

.9250       9661 

.4108       6130 

2.4342       3864 

40 

30 

.  3827       5828 

.9239       9650 

.4142        6172 

2.4142       3828 

30 

40 

.  3854        58.59 

.9228       9651 

.4176       6208 

2.3945       3792 

20 

50 

.3881        5889 

.9216       9646 

.4210       6243 

2  .  3750       3757 

10 

23°  00' 

.3907  9.5919 

.9205  9.9640 

.4245  9.6279 

2.3559  0.3721 

67°  00' 

10 

.3934        5948 

.9194       9635 

.4279       6314 

2  .  3369       3686 

50 

20 

.3961       5978 

.9182       9629 

.4314        6348 

2.3183       3652 

40 

30 

.  3987       6007 

.9171       9624 

.4348       6383 

2.2998       3617 

30 

40 

.4014       6036 

.9159       9618 

.4383       6417 

2.2817       3583 

20 

50 

.4041        6065 

.9147       9613 

.4417       6452 

2.2637       3548 

10 

24°  00' 

.4067  9.6093 

.9135  9.9607 

.4452  9.6486 

2.2460  0.3514 

66°  00' 

10 

.4094       6121 

.9124       9602 

.4487       6520 

2  .  2286       3480 

50 

20 

.4120       6149 

.9112       9596 

.4522       6553 

2.2113       3447 

40 

30 

.4147        6177 

.9100       9590 

.4557       6587 

2.1943       3413 

30 

40 

.4173       6205 

.0088      9584 

.4592       66?0 

2.1775       3380 

20 

50 

.4200       6232 

.  9075       9579 

.4628       6654 

2.1609       3346 

10 

25°  00' 

.4226  9.6259 

.9063  0.9573 

.4663  9.6087 

2.1445  0.3313 

65°  00' 

10 

.4253       6286 

.9051       9567 

.4699        6720 

2.1283       3280 

60 

20 

.4279        6313 

.9038       9561 

.4734       6752 

2.1123       3248 

40 

30 

.4305        6340 

.9026       9555 

.4770       6785 

2.0965       3215 

30 

40 

.4331        6360 

.9013       9549 

.480(5        6817 

2.0809       3183 

20 

50 

.  4358       6392 

.9001        9543 

.4841        6850 

2.0655       3150 

10 

26°  00* 

.4384  9.6418 

.8988  9.9537 

.4877  9.6882 

2.0503  0.3118 

64°  00' 

10 

.4410       6444 

.8975       9530 

.4913       6914 

2.0353       3086 

60 

20 

.4430       6470 

.8962       9524 

.4950       6946 

2.0204       3054 

40 

30 

.4462       6495 

.8040       9518 

.4980       6977 

2.0057       3023 

30' 

40 

.4488       6521 

.8936       9512 

.  5022        7009 

1.9912        2991 

20 

50 

.4514        6546 

.8923       9505 

.  5059        7040 

1  .  9708       2960 

10 

27°00/ 

.4540  9.6570 

.8910  9.9499 

.5095  9.7072 

1.9626  0.2928 

63°  00' 

Nat.      Log. 

Nat.      Log. 

Nat.      Log. 

Nat.       Log. 

Angles 

Cosines 

Sines 

Cotangents 

Tangents 

Angles 

TABLES 


487 


TABLE  XI.— TRIGONOMETRIC  FUNCTIONS— Contin 


Angles 

Sines 

Cosines 

Tangents 

Cotangents 

Angles 

Nat.   Log. 

Nat.   Log. 

Nat.   Log. 

Nat.   Log. 

27°  00' 

.4540  9.6570 

.8910  9.9499 

.5095  9.7072 

1.9626  0.2928 

63°  OC' 

10 

.4566   6595 

.8897   9492 

.5132   7103 

1.9486   2897 

50 

20 

.4592   6620 

.8884   9486 

.5169   7134 

1.9347   2866 

40 

30 

.4617   6644 

.8870   9479 

.5206   7165 

1.9210   2835 

30 

40 

.4643   6668 

.8857   9473 

.5243   7196 

1.9074   2804 

20 

50 

.  4669   6692 

.  8843   9466 

.5280   7226 

1.8940   2774 

10 

28°  00' 

.4695  9.6716 

.8829  9.9459 

.5317  9.7257 

1.8807  0.2743 

62°  00' 

10 

.4720   6740 

.8816   9453 

.5354   7287 

1.8676   2713 

50 

20 

.4746   6763 

.8802   9446 

.5392   7317 

1.8546   2683 

40 

30 

.4772   6787 

.8788   9439 

.5430   7348 

1.8418   2652 

30 

40 

.4797   6810 

.8774   9432 

.5467   7378 

1.8291   2622 

20 

50 

.4823   6833 

.8760   9425 

.  5505   7408 

1.8165   2592 

10 

29°  00' 

.4848  9.6856 

.8746  9.9418 

.5543  9.7438 

1.8040  0.2562 

61°  00' 

10 

.4874   6878 

.8732   9411 

.5581   7467 

1.7917   2533 

50 

20 

.4899   6901 

.8718   9404 

.5619   7497 

1.7796   2503 

40 

30 

.4924   6923 

.8704   9397 

.5658   7526 

1.7675   2474 

30 

40 

.4950   6946 

.  8689   9390 

.5696   7556 

1.7556   2444 

20 

50 

.4975   6968 

.8675   9383 

.5735   7585 

1.7437   2415 

10 

30°  00' 

.5000  9.6990 

.8660  9.9375 

.5774  9.7614 

1.7321  0.2386 

60°  00' 

10 

.5025   7012 

.8646   9368 

.5812   7644 

1.7205   2356 

50 

20 

.5050   7033 

.8631   9361 

.5851   7673 

1.7090   2327 

40 

30 

.  5075   7055 

.8616   9353 

.  5890   7701 

1.6977   2299 

30 

40 

.5100   7076 

.8601   9346 

.5930   7730 

1  .  6864   2270 

20 

50 

.5125   7097 

.8587   9338 

.  5969   7759 

1.6753   2241 

10 

31°  00' 

.5150  9.7118 

.8572  9.9331 

.6009  9.7788 

1.6643  0.2212 

59°  00' 

10 

.5175   7139 

.8557   9323 

.6048   7816 

1.6534   2184 

50 

20 

.5200   7160 

.8542   9315 

.6088   7845 

1.6426   2155 

40 

30 

.5225   7181 

.8526   9308 

.6128   7873 

1.6319   2127 

30 

40 

.5250   7201 

.8511   9300 

.6168   7902 

1.6212   2098 

20 

50 

.5275   7222 

.8496   9292 

.  6208   7930 

1.6107   2070 

10 

32°  00' 

.5299  9.7242 

.8480  9.9284 

.6249  9.7958 

1.6003  0.2042 

58°  00' 

10 

.5324   7262 

.8465   9276 

.6289   7986 

1.5900   2014 

50 

20 

.5348   7282 

.8450   9268 

.6330   8014 

1.5798   1986 

40 

30 

.5373   7302 

.  8434   9260 

.6371   8042 

1.5697   1958 

30 

40 

.5398   7322 

.8418   9252 

.6412   8070 

1.5597   1930 

20 

50 

.5422   7342 

.  8403   9244 

.  6453   8097 

1  .  5497   1903 

10 

33°  00' 

.5446  9.7361 

.8387  9.9236 

.6494  9.8125 

1.5399  0.1875 

57°  00' 

10 

.5471   7380 

.8371   9228 

.6536   8153 

1.5301   1847 

50 

20 

.5495   7400 

.8355   9219 

.6577   8180 

1.5204   1820 

40 

30 

.5519   7419 

.8339   9211 

.6619   8208 

1.5108   1792 

30 

40 

.  5544   7438 

.8323   9203 

.6661   8235 

1.5013   1765 

20 

50 

.  5568   7457 

.8307   9194 

.6703   8263 

1.4919   1737 

10 

34°  00' 

.5592  9.7476 

.8290  9.9186 

.6745  9.8290 

1.4826  0.1710 

56°  00' 

10 

.5616   7494 

.8274   9177 

.6787   8317 

1.4733   1683 

50 

20 

.5640   7513 

.8258   9169 

.6830   8344 

1.4641   1656 

40 

30 

.5664   7531 

.8241   9160 

.6873   8371 

1.4550   1629 

30 

40 

.5688   7550 

.8225   9151 

.6916   8398 

1.4460   1602 

20 

50 

.5712   7568 

.8208   9142 

.6959   8425 

1.4370   1575 

10 

36°  00' 

.5736  9.7586 

.8192  9.9134 

.7002  9.8452 

1.4281  0.1548 

55°  00' 

10 

.5760   7604 

.8175   9125 

.7046   8479 

1.4193   1521 

50 

20 

.5783   7622 

.8158   9116 

.  7089   8506 

1.4106   1494 

40 

30 

.5807   7640 

.8141   9107 

.7133   8533 

1.4019   1467 

30 

40 

.5831   7657 

.8124   9098 

.7177   8559 

1.3934   1441 

20 

50 

.  5854   7675 

.  8107   9089 

.7221   8586 

1.3848   1414 

10 

36°  00' 

.5878  9.7692 

.8090  9.9080 

.7265  9.8613 

1.3764  0.1387 

54°  00' 

Nat.   Log. 

Nat.   Log. 

Nat.   Log. 

Nat.   Log. 

Angles 

Cosines 

Sines 

Cotangents 

Tangents 

Angles 

488 


PR  A  CTICAL  MA  TH  KM  A  TICS 


TABLE  XL— TRIGONOMETRIC  FUNCTIONS— Continued 


Angle* 

Sines 

Cosine* 

Tangents 

Cotangents 

Angles 

Nat.      Log. 

Nat.      Log. 

Nat.      Log. 

Nat.       Log. 

36°  00' 

.5878  9.7692 

V'l'HI     '1     'l(l-vl) 

.7265  978018 

1.3764  0.1387 

54°  00' 

10 

.5901       7710 

XIT:{      '."ITU 

.7310       8080 

1.8080       1361 

50 

20 

.  5925       7727 

.  8056       9061 

.  7355      8000 

1  .  3597       1334 

40 

30 

.  5948       7744 

.8039       9052 

.7400      BOM 

1.3514       1308 

30 

40 

.5972       7761 

.8021       9042 

.7445       8718 

1.3432        1282 

20 

50 

.5995       7778 

.8004       9033 

.  7490       8745 

1.3351       1255 

10 

37°  00' 

.6018  9.7795 

.7986  9.9023 

.7536  9.8771 

1.3270  0.1229 

53°  00' 

10 

.6041       7811 

.7000       9014 

.7581       8797 

1.3190       1203 

50 

20 

.0006       7828 

.7951       9004 

.  7627       8824 

1.3111       1176 

40 

30 

.6088       7844 

.7934       8995 

.7673       8850 

1.3032       1150 

30 

40 

.6111       7861 

.7916       8985 

.  7720       8876 

1.2954       1124 

20 

50 

.6134       7877 

.7898       8975 

.  7766       8902 

1.2876       1098 

10 

38°  00' 

.6157  9.7893 

.7880  9.8965 

.7813  9.8928 

1.2799  0.1072 

52°  00' 

10 

.6180       7910 

.7862       8955 

.  7860       8954 

1.2723       1046 

50 

20 

.6202       7926 

.7844       8945 

.7907       8980 

1.2647       1020 

40 

30 

.6225       7941 

.7826       8935 

.7954       9006 

1  .  2572       0994 

30 

40 

.  6248       7957 

.7808       8925 

.8002       9032 

1.2497       0968 

20 

50 

.  6271       7973 

.7790       8915 

.  8050       9058 

1  .  2423       0942 

10 

39°  00' 

.6293  9.7989 

.7771  9.8905 

.8098  9.9084 

1.2349  0.0916 

51°  00' 

10 

.6310       8004 

.7753       8895 

.8146       9110 

1.2276       0890 

60 

20 

.  6338       8020 

.7735       8884 

.8195       9135 

1  .  2203       0865 

40 

30 

.6361       8035 

.7716       8874 

.8243       9161 

1.2131       0839 

30 

40 

.6383       8050 

.7698       8864 

.8292       9187 

1.2059       0813 

20 

50 

.  6406       8066 

.7679       8853 

.8342       9212 

1  .  1988       0788 

10 

40°  00' 

.6428  9.8081 

.7660  9.8843 

.8391  9.9238 

1.1918  0.0762 

50°  00' 

10 

.6450       8096 

.7642       8832 

.8441       9264 

1.1847       0736 

50 

20 

.6472       8111 

.7623       8821 

.  8491       9289 

1.1778       0711 

40 

30 

.6494       8125 

.7604       8810 

.8541       9315 

1.1708       0685 

30 

40 

.6517       8140 

.  7585       8800 

.8591       9341 

1.1640       0659 

20 

50 

.6539       8155 

.  7566       8789 

.8642       9366 

1.1571       0634 

10 

41°  00' 

.6561   9.8169 

.7547  9.8778 

.8693  9.9392 

1.1504  0.0608 

49°  00' 

10 

.6583       8184 

.7528       8767 

.8744       9417 

1  .  1436       0583 

60 

20 

.6604       8198 

.7509       8756 

.8796       9443 

1.1369       0557 

40 

30 

.6626       8213 

.7490       8745 

.  8847       9468 

1.1303       0532 

30 

40 

.  6648       8227 

.7470       8733 

.8899       9494 

1  .  1237       0506 

20 

50 

.  6670       824  1 

.7451        8722 

.8952       9519 

1.1171       0481 

10 

42"  00' 

.6691  9.8255 

.7431  9.8711 

.9004  9.9544 

1.1106  0.0456 

48«00' 

10 

.6713       8269 

.7412       8699 

.9057       9570 

1.1041        0430 

50 

20 

.  6734       8283 

.  7392       8688 

.9110       9595 

1.0977       0405 

40 

30 

.  6756       8297 

.  7373       8676 

.9163       9621 

1.0913       0379 

30 

40 

.6777       8311 

.  7353       8665 

.9217       9646 

1.0850       0354 

20 

50 

.  6799       8324 

.  7333       8653 

.9271       9671 

1.0786       0329 

10 

43°  00' 

.6820  9.8338 

.7314  9.8641 

.9325  9.9697 

1.0724  0.0303 

47°  00' 

10 

.6841       8351 

.  7294       8629 

.9380       9722 

1.0661       0278 

50 

20 

.0862       8365 

.7274       8818 

.9435       9747 

1  .  0599       0253 

40 

30 

.6884       8378 

.  7254       8606 

.9490       9772 

1  .  0538       0228 

30 

40 

.6905       8391 

.  7234       8594 

.9545       9798 

1.0477       0202 

20 

50 

.  6926       8405 

.7214       8582 

.9601       9823 

1.0416       0177 

10 

44°  00' 

.6947  9.8418 

.7193  9.8569 

.9657  9.9848 

1.0355  0.0152 

46°  00' 

10 

.6967       8431 

.7173       8557 

.9713       9874 

1.0295       0120 

60 

20 

.0088       8444 

.7153       8545 

.9770       9899 

1.0235       0101 

40 

30 

.7009       8457 

.7133       8532 

.9827       9924 

1.0176       0070 

30 

40 

.7030       8469 

.7112       8520 

.9884       9949 

1.0117       0051 

20 

50 

.  7050       8482 

.  7092       8507 

.9942       9975 

1  .  0058       0025 

10 

45°  00' 

.7071  9.8495 

.7071  9.8495 

1.0000  0.0000 

1   0000  0.0000 

45°  00' 

Nat.      Log. 

Nat.      Log. 

Nat.      Log. 

Nat.       Log. 

Angles 

Cosines 

Sines 

Cotangents 

Tangents 

Angles 

INDEX 


(Numbers  refer  to  pages.) 


Abscissa,  353 
Absolute  value,  245 
Accuracy  of  results,  43 
Altitude  of  triangle,  113 
Anchor  ring,  226 
Angle,  111,  396 

acute,  112 

central,  148 

complementary,  112 

construction  of,  179,  409 

definition,  111,  396 

finding  from  table,  415 

inscribed,  149 

measurement  of,  177,  179,  398 

negative,  397,  444 

obtuse,  112 

of  depression,  429 

of  elevation,  429 

right,  111 

positive,  397 

supplementary,  112 

units  for  measuring,  177 

vertex  of,  111 
Angles  and  complements,  407,  408, 

442 

Angles  and  supplements,  442 
Angles,  sums  of,  455 
Antecedent,  85 
Approximate  results,  52 
Arc,  147 

Average  ordinate  rule,  364 
Averages,  77 
Axioms,  259 

Beams,  wooden,  346 
Bearing,  430 
Belt  pulleys,  171 
Belts,  carrying  power,  10 
lengths  of,  433 


Binomial,  234 
Bisector,  113 
Board  foot,  122 
Boiler  capacity,  209 

tubes,  209 
Brickwork,  201 

Cancellation,  7 
Characteristics,  374 

rules  for,  374 
Checking,  55 

Chords  of  angles,  table,  478 
Circle,  147 

area  of,  151 

to  find  radius  of,  189 
Circular  motion,  449 
Circumference,  147 
Coefficient,  233 
Completing  the  square,  334 
Cone,  214 

area  of,  216 

frustum  of,  215 

volume  of,  216 
Consequent,  85 
Constant,  340,  347 
Contracted  methods,  52 
Corresponding  sides,  131 
Cosine,  404 
Cosines,  law  of,  460 
Cube,  195 
Curve,  cycle  of,  450 
Curves,  cosecant,  448 

cosine,  448 

cotangent,  448 

degree  of,  401 

frequency  of,  452 

periodic,  449 

railroad,  401 

sine,  447 
489 


490 


INDEX 


Curves,  tangent,  448 
Cutting  speeds,  170 
Cylinder,  203 

area  of,  203 

hollow,  204 

inscribed,  203 

volume  of,  204 

Decagon,  189 
Denominator,  12 

least  common,  16,  295 
Density,  94 
Diagonal,  113 
Diameter,  147 
Discounts,  78 
Divisors,  rules  for,  3 
Drawing  to  scale,  181 

Elimination,  320 

Ellipse,  155 

Emery  wheels,  speeds  of,  160,  402 

Equations,  257,  332 

conditional,  257 

equivalent,  320 

identical,  257 

inconsistent,  320 

independent,  320 

indeterminate,  319 

quadratic,  287,  332 

simultaneous,  319 

solutions  of,  258 
Exponents,  98,  233,  327 

laws  of,  268,  327,  370 
Extremes,  86 

Factor,  2 

greatest  common,  3 

of  safety,  348 

Feed  of  cutting  tool,  33,  170 
Flywheel,  velocity  of,  160 
Foot-pounds,  312 
Formulas,  238,  290,  470 

difference  of  angles,  455 

products  to  sums,  457 

sums  of  angles,  455 

sums  to  products,  457 

table  of.  470 


Fractions,  12,  293 

decimal,  38 

improper,  12 

proper,  12 

principles  of,  14 
Frequency  factor,  453 
Frustum,  215 

area  of,  216 

volume  of,  217 

Gear  wheels,  171 

Grade,  87 

Graphs,  351 

Grindstones,   speeds  of,    161,   402 

Heights,  measuring,  89,  134 
Helix,  439 
Heptagon,  188 
Hexagon,  113,  188 

regular,  141 
Horizon,  dip  of,  436 

distance  of,  435 
Horse-power,  10,  312 

coal  for,  47 

effective,  365 

indicated,  365 

of  waterfall,  393 

transmitted  by  belt,  393 
Hydraulic  machines,  93 
Hypotenuse,  126 

Indebtedness  paid  in  installments, 

390 

Integer,  2 
Interest,  83 
Interpolating,  356 
Interpolation,  377 
Inverse  functions,  426 
Involution,  98 

Kilogram,  64 

Law  of  cosines,  460 

of  sines,  459 
Lever,  91 
Line,  110 

broken,  110 


INDEX 


491 


Line,  curved,  110 

division  into  parts,  186 

horizontal,  112 

straight,  110 

vertical,  112 
Lines,  parallel,  111 
Liter,  64 
Logarithms,  definition,  370 

history  of,  370 

hyperbolic,  384 

Napierian,  384 

natural,  384 
Lumber,  121 

Machine  screw  proportions,  49 
Mantissa,  definition^  374 

rules  for,  377 
Mariner's  compass,  430 
Means,  86 
Measures,  English,  58 

metric,  62 
Median,  113 
Meter,  63 

Micrometer,  32,  192 
Mil,  circular,  175 

square,  176 
Mixed  number,  12 
Monomial,  234 
Multiple,  3 

least  common,  3,  295 

Number,    corresponding    to    loga- 
rithm, 379 

rules  for,  380 
Numbers,  algebraic,  245 

composite,  3 

definite,  231 

even,  2 

general,  231 

negative,  244 

odd,  3 

positive,  244 

prime,  3 

relative,  244 

useful,  table  of,  474 
Numerator,  12 


Octagon,  113,  187,  189 
Ohm's  Law,  316 
Ordinate,  353 

Parallelogram,  114 

area  of,  115 
Pentagon,  113,  188 
Percentage,  74 

cases  of,  74 
Per  cent  of  error,  77 
Perimeter,  112 
Plotting,  354 
Point,  110 
Polygon,  112 

area  of,  115 

circumscribed,  148 

inscribed,  148 

regular,  113 
Polynomial,  234 
Power,  98,  233 
Powers  by  logarithms,  383 
Prism,  194 

surface  of,  195 

volume  of,  196 
Prismatoid,  227 
Products,  approximate,  273 

by  logarithms,  381 

representation  of,  272 
Projection,  orthogonal,  427 
Proportion,  86 
Protractor,  178 
Pulleys,  rules  for,  9 
Pyramid,  214 

area  of,  216 

volume  of,  217 

Quadrants,  353,  398 
Quadrilateral,  113 
Quotients  by  logarithms,  381 

Radian,  177,  398 
Radius,  147 
Rafters,  135 
Railroad  curves,  400 
Ratio,  85 

inverse,  85 


402 


INDEX 


Rectangle,  114 

area  of,  115 

Resultant*,  21,  466,  468 
Right  triangles,  solution  of,  421 
Ring,  area  of,  153 
Root,  square,  99 

rule  for,  105 

Roots  by  logarithms,  383 
Roots  of  an  equation,  288 


Safety  valve,  163 
Screw  gearing,  33 
Screw  threads,  31,  143,  437 
Sector,  148 

area  of,  153 
Segment,  147 

area  of,  154,  434 
Sheet  metal  gage  table,  479 
Shingles,  122 
Signs,  grouping,  232 

operation,  245 

quality,  245 

Similar  figures,  106,  130 
Simpson's  Rule,  363 
Sine,  404 
Sines,  law  of,  459 
Slide  rule,  237 
Solid,  geometric,  109 

rectangular,  195 
Specific  gravity,  94 

table  of,  480 
Speeds,  emery  wheels,  160,  402 

grindstones,  161,  402 

polishing  wheels,  402 
Sphere,  220 

area  of,  220 

segment  of,  221 

volume  of,  221 

zone  of,  222 
Spirals,  439 

lead  of,  439 
Square,  114 
Squared  paper,  351 
Steam  indicator  diagram,  365 
Steel  square,  134 

uses  of,  131,  159 


Stonework,  200 

Strength  of  materials,  table,  481 

Surfaces,  109 

curved,  110 

plane,  110 


Tabie  of  formulas,  470 

of  logarithms,  482 

of  functions,  484 
Tabular  difference,  376 
Tangent,  404 
Tapers,  131 
Testa,  251,  260,  270 
Thermometers,  311 
Threads,  acme,  437 

sharp- V,  144 

U.  S.  S.,  144 

Whitworth,  437 
Trapezoid,  114 

area  of,  118 
Triangle,  113 

area  of,  116,  117 

construction  of,  181 

equilateral,  139 

isosceles,  139 

oblique,  459 

right,  126,  421 
Trigonometric  functions,  404 

computation  of,  410 

definitions  of,  404 

finding  by  tables,  414 

inverse,  426 

of  difference  of  angles,  4? 

of  half  an  angle,  456 

of  sum  of  angles,  455 

of  twice  an  angle,  456 

table  of,  484 

Trigonometry,  uses  of,  396 
Trinomial,  234 
Turning,  132 
Twist  drill,  34 

Useful  number*1,  474 

Variable,  340 
Variation,  340 


INDEX  493 

Variation,  direct,  340  Water  flow  in  pipes,  394 

indirect,  342  Weights  of  round  iron,  34 

joint,  343  Weights,  tables  of,  59 

Vectors,  427,  467  Wire  gages,  tables,  479 

Vernier,  191  Worm  wheel,  34 

Vertex,  111 

Voids,  80  Zones  of  sphere,  222 


UNIVERSITY  OF  CALIFORNIA  LIBRARY,  LOS  ANGELES 

COLLEGE  LIBRARY 

This  book  is  due  on  the  last  date  stamped  below. 


•an    ' 


I96' 


1968 

CDC.  DB. 


OCT161969 


Book  Slip-25m-7,'61(Cl437b4)4280 


College 
Library 


UCLA-College  Library 
QA39P18 


L  005  738  177  4 


JT>  I/BRANCH, 


>S 


mxS 


